Final answer:
Robbie Knievel's launch angle for his jump across the Grand Canyon was approximately -5.35 degrees.
Explanation:
To calculate the launch angle of Robbie Knievel's jump across the Grand Canyon, we can use the horizontal distance and initial velocity of the motorcycle. The horizontal distance is given as the width of the canyon, which is 65 meters. The initial velocity is given as 36 m/s. We can use the equation for horizontal motion to find the time of flight:
d = v0xt
Where d is the horizontal distance, v0x is the initial velocity in the x-direction, and t is the time of flight. Solving for t, we get:
t = d / v0x
Substituting the given values, we have:
t = 65 m / 36 m/s = 1.81 s
Next, we can use the equation for vertical motion to find the launch angle:
y = v0yt + (1/2)gt2
Where y is the vertical distance, v0y is the initial velocity in the y-direction, g is the acceleration due to gravity, and t is the time of flight. The vertical distance is zero since Robbie reached the other side of the canyon. Solving for v0y and using the given value of t, we get:
v0y = - (1/2)gt
v0y = - (1/2)(3.7 m/s2)(1.81 s) = -3.34 m/s
Finally, we can use the launch angle formula:
tan(θ) = v0y / v0x
Substituting the calculated values, we have:
tan(θ) = -3.34 m/s / 36 m/s
θ = atan(-3.34 m/s / 36 m/s) = -5.35 degrees
Therefore, Robbie Knievel's launch angle was approximately -5.35 degrees.
Free public education is mandated from: A. Birth B. Preschool through grade 12 for all children C. Kindergarten through grade 12 for all children D. First grade through grades 12 in all states E. Preschool through grade 12 for special needs children
Answer:
C.) First grade through grades 12 in all states
Explanation:
Free public education is mandated from Kindergarten through grade 12 for all children. The correct option is option (c).
Free public education is mandated for all children in the United States from kindergarten through grade 12. This means that public schools are required to provide education to children starting from kindergarten up to the completion of 12th grade. This mandate ensures that all children have access to education without financial barriers during these years.
While preschool education is not universally mandated, some states or districts may offer free or subsidized preschool programs for certain age groups or based on specific criteria.
Special needs children, as mentioned in option E, also fall under the mandate for free public education from preschool through grade 12. Special education services are provided to ensure that children with disabilities receive an appropriate education tailored to their individual needs.
Therefore, option C, "Kindergarten through grade 12 for all children," best represents the mandate for free public education in the United States.
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Consider two cylinders of gas identical in all respects except that one contains O₂ and the other He. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater pressure increase?
A)It's impossible to tell from the information given.
B)the oxygen (O₂)
C)Neither; both will show the same increase.
D)the He
Answer:
D) He
Explanation:
For adiabatic change
[tex]PV^\gamma = constant[/tex]
\frac{P_2}{P_1} =(\frac{V_1}{V_2} )^\gamma
[tex]\frac{P_2}{P_1} =(\frac{3}{1} )^\gamma[/tex]
[tex]P_2 = P_1 3^\gamma[/tex]
As the value of \gamma increases, P_2 increases.
\gamma for He is 1.67 and \gamma for O₂ IS 1.4
Naturally The value of P_2 is greater for He. Increase in pressure for He is greater.
You drive a car 2.0 h at 40 km/h, then 2.0 h at 80 km/h. (a) What is your average velocity? km/h (b) Do you get the same answer if you drive 120 km at each of the two speeds above? no yes
Answer:
a)(2×40+2×80)/(2+2)=240/4=60
b)(240)/(120/40+120/80)=240/(3+1.5)
240/4.5=53.33333
no
Use vectors to decide whether the triangle with vertices P(1, −2, −1), Q(2, 1, −3), and R(6, −1, −4) is right-angled. Yes, it is right-angled. No, it is not right-angled.
Answer:
Yes, it is right-angled
Explanation:
Two vectors are orthogonal if the scalar product between them is zero. Then, we will match each pair of vertices with a vector, wich is formed with the following formula:
Given two points A and B, the vector going from A to B is
[tex]AB=B-A=(B_{x}-A_{x},B_{y}-A_{y},B_{z} -A_{z})[/tex]
So, we calculate each component separately.
[tex]PQ=Q-P=(2-1,1-(-2),-3-(-1))=(1,3,-2)[/tex]
[tex]QR=R-Q=(6-2,-1-1,-4-(-3))=(4,-2,-1)[/tex]
[tex]RP=P-R=(1-6,-2-(-1),-1-(-4))=(-5,-1,3)[/tex]
Finally, using the scalar product formula
[tex]A*B=A_{x}* B_{x}+ A_{y}* B_{y}+ A_{z}* B_{z}[/tex]
we see if the products is zero
[tex]PQ*QR=1*4+3(-2)+(-2)*(-1)=0[/tex]
In this case we don't even have to calculate the other products as we've found that PQ and QR form a right angle.
Final answer:
By calculating the dot product of vectors representing the sides of the triangle, it is found that the dot product of PQ and QR is zero, indicating that these vectors are orthogonal. Therefore, the triangle PQR is right-angled.
Explanation:
To determine whether the triangle with vertices P(1, -2, -1), Q(2, 1, -3), and R(6, -1, -4) is right-angled, we can use the concept of the dot product between two vectors. A triangle is right-angled if one of the pairs of vectors corresponding to its sides is orthogonal, meaning their dot product is zero.
First, we find the vectors representing the sides of the triangle:
⇨ PQ = Q - P = (2, 1, -3) - (1, -2, -1) = (1, 3, -2)
⇨ PR = R - P = (6, -1, -4) - (1, -2, -1) = (5, 1, -3)
⇨ QR = R - Q = (6, -1, -4) - (2, 1, -3) = (4, -2, -1)
Next, calculate the dot products of each pair of vectors:
PQ ⋅ PR = (1, 3, -2) ⋅ (5, 1, -3) = 1*5 + 3*1 - 2*(-3) = 5 + 3 + 6 = 14
PQ ⋅ QR = (1, 3, -2) ⋅ (4, -2, -1) = 1*4 + 3*(-2) - 2*(-1) = 4 - 6 + 2 = 0
PR ⋅ QR = (5, 1, -3) ⋅ (4, -2, -1) = 5*4 + 1*(-2) - 3*(-1) = 20 - 2 + 3 = 21
As we see, the dot product of PQ and QR is zero, which means these two vectors are orthogonal, and hence the triangle PQR is right-angled.
A race car driver on the Flats first heads north for 6.71 km, then makes a sharp turn and heads southwest for 1.33 km, then makes another turn and heads east for 3.67 km. How far is she from where she started?
Answer:
D=7.385km
Explanation:
First we calculate how much it moves vertically (north and south), we will call this distance Y,.
taking into account that the car moves to the southwest and that it has a slope of 45 degrees, we can say that the component that moved south can be subtracted from the distance that moved north.
Y=6.71-1.33sen45=5.77km
then the distance that moves in the east we will call it X
we can add the component of the distance that moves to the southwest with the distance that moves to the east
X=3.67+1.33cos45=4.61Km
then we use the distance equation
[tex]D=\sqrt{x^{2} +y^{2} }[/tex]
[tex]D=\sqrt{4.61^{2}+5.77^{2} }[/tex]
D=7.385km
Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle orbiting in Saturn's D Ring, whose distance from Saturn is 69000 km. How many times larger is the velocity of the particle in the D Ring compared to the particle in the Encke Division?
Answer:
The particle in the D ring is 1399 times faster than the particle in the Encke Division.
Explanation:
The circular velocity is define as:
[tex]v = \frac{2 \pi r}{T}[/tex]
Where r is the radius of the trajectory and T is the orbital period
To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:
[tex]T^{2} = r^{3}[/tex]
Where T is orbital period and r is the radius of the trajectory.
Case for the particle in the Encke Division:
[tex]T^{2} = r^{3}[/tex]
[tex]T = \sqrt{(133370 Km)^{3}}[/tex]
[tex]T = \sqrt{(2.372x10^{15} Km)}[/tex]
[tex]T = 4.870x10^{7} Km[/tex]
It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( [tex]1.50x10^{8} Km[/tex] )
1 AU is defined as the distance between the earth and the sun.
[tex]\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU[/tex]
[tex]T = 0.324 AU[/tex]
But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.
[tex]T = \frac{0.324 AU}{1 AU} . 1 year[/tex]
[tex]T = 0.324 year[/tex]
That can be expressed in units of days
[tex]T = \frac{0.324 year}{1 year} . 365.25 days[/tex]
[tex]T = 118.60 days[/tex]
Circular velocity for the particle in the Encke Division:
[tex]v = \frac{2 \pi r}{T}[/tex]
[tex]v = \frac{2 \pi (133370 Km)}{(118.60 days)}[/tex]
For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.
[tex]118.60 days .\frac{86400 s}{1 day}[/tex] ⇒ 10247040 s
[tex]133370 Km .\frac{1000 m}{1 Km}[/tex] ⇒ 133370000 m
[tex]v = \frac{2 \pi (133370000 m)}{(10247040 s)}[/tex]
[tex]v = 81.778 m/s[/tex]
Case for the particle in the D Ring:
For the case of the particle in the D Ring, the same approach used above can be followed
[tex]T^{2} = r^{3}[/tex]
[tex]T = \sqrt{(69000 Km)^{3}}[/tex]
[tex]T = \sqrt{(3.285x10^{14} Km)}[/tex]
[tex]T = 1.812x10^{7} Km[/tex]
[tex]\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU[/tex]
[tex]T = 0.120 AU[/tex]
[tex]T = \frac{0.120 AU}{1 AU} . 1 year[/tex]
[tex]T = 0.120 year[/tex]
[tex]T = \frac{0.120 year}{1 year} . 365.25 days[/tex]
[tex]T = 43.83 days[/tex]
Circular velocity for the particle in D Ring:
[tex]v = \frac{2 \pi r}{T}[/tex]
[tex]v = \frac{2 \pi (69000 Km)}{(43.83 days)}[/tex]
For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.
[tex]43.83 days . \frac{86400 s}{1 day}[/tex] ⇒ 3786912 s
[tex]69000 Km . \frac{1000 m}{ 1 Km}[/tex] ⇒ 69000000 m
[tex]v = \frac{2 \pi (69000000 m)}{(3786912 s)}[/tex]
[tex]v = 114.483 m/s[/tex]
[tex]\frac{114.483 m/s}{81.778 m/s} = 1.399[/tex]
The particle in the D ring is 1399 times faster than the particle in the Encke Division.
A 230.-mL sample of a 0.240 M solution is left on a hot plate overnight; the following morning, the solution is 1.75 M. What volume of water evaporated from the 0.24 M solution?
Answer:
The volume of water evaporated is 199mL
Explanation:
Concentration is calculated with the following formula
[tex]C=\frac{n}{V}[/tex]
where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.
So we isolate the variable n to know the amount of moles, using the volume given in liters
[tex]230mL=0.23L[/tex]
[tex]n=C*V=0.240 M*0.23L=0.055 mol[/tex]
Now, we isolate the variable V to know the new volume with the new concentration given.
[tex]V=\frac{n}{C} =0.055mol/1.75M=0.031L=31mL[/tex]
Finally, the volume of water evaporated is the difference between initial and final volume.
[tex]V_{ev}= V_{i} -V_{f} =230mL-31mL=199mL[/tex]
A child bounces a 52 g superball on the sidewalk. The velocity change of the superball is from 20 m/s downward to 14 m/s upward. If the contact time with the sidewalk is 1 800 s, what is the magnitude of the average force exerted on the superball by the sidewalk? Answer in units of N.
Answer:
0.982 N
Explanation:
mass of ball, m = 52 g = 0.052 kg
initial velocity, u = - 20 m/s (downward)
final velocity, v = 14 m/s (upward)
time of contact, t = 1.8 s
According to Newton's second law, the rate of change of momentum of the body is equal to the forced exerted on that body.
initial momentum , pi = mass x initial velocity = 0.052 x (-20) = - 1.04 kg m/s
final momentum, pf = mass x final velocity = 0.052 x 14 = 0.728 kg m/s
Change in momentum = final momentum - initial momentum
= 0.728 - (- 1.04)= 1.768 kg m/s
So, force = change in momentum / time
Force = 1.768 / 1.8 = 0.982 N
The magnitude of the average force exerted on the superball by the sidewalk is 1414.4 N. This is calculated by determining the impulse based on the change in velocity and mass of the superball, and dividing the impulse by the contact time.
Explanation:The question asks to calculate the magnitude of the average force exerted on a superball by the sidewalk during its bounce. The superball's mass is 52 g (0.052 kg), and it experiences a velocity change from 20 m/s downward to 14 m/s upward. The contact time with the sidewalk is given as 1/800 s. First, we must calculate the change in velocity (impulse) and then use it to find the average force.
To calculate the impulse, we add the magnitudes of both components of velocity together since the ball changes direction, the total change in velocity (Δv) is 20 m/s (downward) + 14 m/s (upward) = 34 m/s. The impulse (J) is then given by Δp = mΔv, where m is the mass of the ball. Therefore, J = 0.052 kg * 34 m/s = 1.768 kg·m/s.
The average force (Favg) can be found by dividing the impulse by the contact time (Δt): Favg = J / Δt. Thus, Favg = 1.768 kg·m/s / (1/800 s), which equals an average force of 1414.4 N.
To qualify to run in the 2005 Boston Marathon, a distance of 26.2 miles, an 18-year-old woman had to have completed another marathon in 3 hours and 40 minutes or less. a) To qualify, what must a woman's average speed have been in miles per hour?b) To qualify, what must a woman's average speed have been in meters per second? There are 1609 m in a mile.
Answer:
7,14545 mph and 3,1936 m/s
Explanation:
The average speed is calculated by dividing the displacement over time, then it is 26,2 miles/(3 2/3 hours), here 3 (2/3) hours is a mixed number, that represents 11/3 hours or 3,66 hours. Then the average speed is 7,14545 mph, now to turn this into meters per second, we notice as mentioned that 1 mile =1609 meters and 1 hour=3600 seconds. Then 7,14545 miles/hour* (1 hour/3600 seconds) * (1609 meters/1 mile)=3,1936 m/s
A plane flying from Townsville, Australia, has an air speed of 264 m/s in a direction 5.0° south of west. It is in the jet stream, which is blowing at 37 m/s in a direction 15° south of east. In this problem you are going to be asked about the velocity of the airplane relative to the Earth.
Answer:
Speed of plane is 300.5 m/s at angle of 6.22 degree South of West
Explanation:
Air speed of the plane is given as
v = 264 m/s in direction 5 degree South of West
So we have
[tex]v_1 = 264 cos5 \hat i + 264 sin5 \hat j[/tex]
[tex]v_1 = 263 \hat i + 23 \hat j[/tex]
Also we have speed of air is given as
v = 37 m/s at 15 degree South of West
so it is
[tex]v_2 = 37 cos15\hat i + 37 sin15 \hat j[/tex]
[tex]v_2 = 35.74 \hat i + 9.58 \hat j[/tex]
So the net speed of plane with respect to ground is given as
[tex]v_p = v_1 + v_2[/tex]
[tex]v_p = (263 \hat i + 23 \hat j) + (35.74 \hat i + 9.58 \hat j)[/tex]
[tex]v_p = 298.74 \hat i + 32.58\hat j[/tex]
so it is
[tex]v_p = \sqrt{298.74^2 + 32.58^2}[/tex]
[tex]v_p = 300.5 m/s[/tex]
direction is given as
[tex]\theta =tan^{-1} \frac{v_y}{v_x}[/tex]
[tex]\theta = tan^{-1} \frac{32.58}{298.74}[/tex]
[tex]\theta = 6.22 degree[/tex]
The froghopper Philaenus spumarius is supposedly the best jumper in the animal kingdom. To start a jump, this insect can accelerate at 4.00 km/s2 over a distance of 2.0 mm as it straightens its specially designed "jumping legs."
(a) Find the upward velocity with which the insect takes off.
(b) In what time interval does it reach this velocity?
(c) How high would the insect jump if air resistance were negligible? The actual height it reaches is about 70 cm, so air resistance must be a noticeable force on the leaping froghopper.
Answer:
Part a)
[tex]v_f = 4 m/s[/tex]
Part b)
[tex]t = 0.001 s[/tex]
Part c)
[tex]d = 0.815 m[/tex]
Explanation:
Part a)
As we know that initially the grass hopper is at rest at the ground position
Now the acceleration is given as
[tex]a = 4000 m/s^2[/tex]
distance of the legs that it stretched is given as
[tex]s = 2.0 mm[/tex]
so we have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]v_f^2 - 0 = 2(4000)(0.002)[/tex]
[tex]v_f = 4 m/s[/tex]
Part b)
time taken to reach this speed is given as
[tex]v_f - v_i = at[/tex]
[tex]4 - 0 = 4000 t[/tex]
[tex]t = 0.001 s[/tex]
Part c)
as the grass hopper reach the maximum height its final speed would be zero
so we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex]0 - 4^2 = 2(-9.81) d[/tex]
[tex]d = 0.815 m[/tex]
Final answer:
The froghopper Philaenus spumarius takes off at an upward velocity of 4 m/s and reaches this velocity in 0.001 seconds. In the absence of air resistance, it can potentially jump to a height of 0.815 m, while air resistance reduces its actual jump to about 70 cm.
Explanation:
The froghopper Philaenus spumarius, known for its powerful jumping ability, can accelerate at 4.00 km/s² over a distance of 2.0 mm. To find the upward velocity (a) with which the insect takes off, the following kinematic equation can be used:v² = u² + 2as
where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration, and s is the distance.
Firstly, convert the given units to make them consistent. The acceleration becomes 4000 m/s² and the distance becomes 0.002 m. Plugging the values into the equation gives:
v² = 0 + 2(4000)(0.002) = 16
The takeoff velocity, v, is √16, which equals 4 m/s.
For the time interval (b), we use the equation:
v = u + at
So, t = v / a = 4 / 4000 = 0.001 s.
To calculate how high the insect would jump if air resistance were negligible (c), we can use the equation:
s = ut + ½at²
Since initial velocity u is 0 when coming back down, and final velocity v is 0 at the peak of the jump, we set v = 0 and solve for s:
0 = (4)2 - 2(9.81)s
s = 16 / 19.62 = 0.815 m (or 81.5 cm).
Comparing this to the actual jump height of 70 cm indicates a significant effect of air resistance on the insect's jump.
A baseball pitcher throws a baseball with a speed of 42 m/s . In throwing the baseball, the pitcher accelerates it through a displacement of about 3.5 m, from behind the body to the point where it is released.
Answer:
a = 252 m / s²
Explanation:
The initial speed Vο = 0
The speed of baseball with witch it is thrown Vf = 42 m/s
Displacement covered = 3.5 m
Hence acceleration will be constant at this moment, so as per the equation of constant acceleration : ( Vf² - Vο² ) / 2d = a
a = 252 m / s²
A rock is dropped from a tall building. If the rock starts at rest , and the origin of the coordinate system is taken at the pinnacle of the building with positive direction taken to be upward direction, after 5.18 seconds:
a) What is the displacement of the rock?
b) What is the velocity of the rock?
Answer:
a)y=-131.47 m :if the coordinate system is taken at the pinnacle of the building with positive direction taken to be upward direction.
b)v=-50.764m/s:The minus sign indicates the direction of the speed that is down
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
y = v₀*t +½ g*t² Equation 1
v=v₀+g*t Equation 2
y: The vertical distance the ball moves at time t
v₀: Initial speed in m/s
g= acceleration due to gravity in m/s²
v= Speed the ball moves at time t
Known information
v₀=0
t=5.18 s
g=9.8 m/s²
Development of problem
a)We replace t in the equations (1) to y(5.18s) :
y =o+½ *9.8*5.18² =131.47m
y=-131.47 m :if the coordinate system is taken at the pinnacle of the building with positive direction taken to be upward direction
b)We replace t in the equations (2) to v(5.18s) :
v=o+9.8*5.18=50.764m/s
v=-50.764m/s:The minus sign indicates the direction of the speed that is down
What is the approximate mass of a column of air 1 cm2 in area that extends from sea level to the upper atmosphere? What is the weight of this amount of air? What is the pressure at the bottom?
Answer:
(a) the approximate mass of a column of air is: 1225 (Kg), (b) the weight of this amount of air is: 12017.25 (N) and (c) the pressure at the bottom is: 120172.5 (KPa)
Explanation:
We need to remember that the equation that related mass and volume is the density as:[tex]p=\frac{m}{V}[/tex], where p is the density, m is the mass and V is the volume and assuming that air density is 1.225 (Kg/m3) and knowing that exosphere is the upper atmosphere and its height is 10000 (Km). We now can get the volume as:[tex]Volume=A*h_{upper atmosphere} =0.0001*10000000=1000(metre^{3})[/tex], so that the approximate mass of a column of air is: [tex]m=p*V=1.225*1000=1225 (Kg)[/tex]. Then the weight of this amount of air is:[tex]W=m*g[/tex], where W is the weight, m is the mass and g the gravity acceleration, so:[tex]W=1225*9.81=12017.25 (N)[/tex]. Finally to find the pressure we know that gauge pressure equation [tex]P=p*g*h[/tex] where P is the pressure, p is the density of the column and h is the height of the column, so [tex]P=1.225*9.81*10000000=120172.5 (KPa)[/tex].
The approximate mass of the air column can be calculated using the density of air and the volume of the column. The weight of the air column can be calculated by multiplying the mass by the acceleration due to gravity. The pressure at the bottom can be calculated by dividing the weight by the area.
Explanation:The approximate mass of a column of air with an area of 1cm² that extends from sea level to the upper atmosphere can be calculated using the density of air. The density of air near the Earth's surface is about 1.29 kg/m³. Using this density, we can calculate the mass by multiplying the density with the volume of the column of air. The volume can be calculated by multiplying the area of 1 cm² with the height of the column, which is the distance from sea level to the upper atmosphere.
To calculate the weight of this amount of air, we need to multiply the mass of the air column by the acceleration due to gravity. The acceleration due to gravity is approximately 9.81 m/s². Multiplying the mass by the acceleration due to gravity will give us the weight of the air column in newtons.
The pressure at the bottom, or at sea level, can be calculated using the weight of the air column and the area of 1 cm². The pressure is equal to the weight divided by the area.
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An electrical short cuts off all power to a submersible diving vehicle when it is a distance of 28 m below the surface of the ocean. The crew must push out a hatch of area 0.70 m2 and weight 200 N on the bottom to escape.If the pressure inside is 1.0 atm, what downward force must the crew exert on the hatch to open it?
To open the hatch of a submerged vehicle, the crew needs to overcome the difference in pressure applied on both sides of the hatch. By calculating the external pressure at the depth, the force outside is found. Subtracting the force inside (internal pressure plus hatch weight) gives the requisite force (~1.95 x 10⁵ N).
Explanation:The question is about determining the force the crew must exert to open the hatch of a submerged vehicle. In physics, this problem involves understanding the principles of fluid statics and the pressure difference between the interior and exterior of the hatch, contributing to the net force on the hatch.
Given that the vehicle is submerged 28 meters in saltwater, with an internal pressure of 1 atm, the external pressure at that depth will be greater than the internal pressure. The pressure a diver experiences under water increases by 1 ATA every 33 feet of salt water. Hence, at 28 meters (approximately 92 feet), the external pressure is about 1 (atmospheric pressure) + 92/33 (pressure due to ocean) = ~3.8 ATA. This is the pressure acting against the crew from outside.
The force acting on the hatch from outside (pressurized water) can be obtained by multiplying the pressure with the area (Force = Pressure x Area). So the Force outside = 3.8 atm x 0.70 m² = ~2.66 x 10⁵ N (considering that 1 atm is approximately 1.013 x 10⁵ N/m²).
Now, the force acting from inside consists of the atmospheric pressure (1 atm) and the weight of the hatch. Force inside = (1 atm x 0.70 m²) + weight of the hatch = 0.71 x 10⁵ N + 200 N = ~0.71 x 10⁵ N
Therefore, the resultant force (force required to open the hatch) is the difference between the outside and inside forces. That becomes ~2.66 x 10⁵ N - 0.71 x 10⁵ N = ~1.95 x 10⁵ N.
This is the downward force the crew must exert on the hatch to open it and escape.
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The downward force that the crew must exert on the hatch to open it at a depth of 28 m below the surface of the ocean is [tex]\( {199567.2 \, \text{N}} \)[/tex].
To find the downward force (F) that the crew must exert on the hatch to open it at a depth of 28 m below the surface of the ocean, we need to consider the hydrostatic pressure acting on the hatch.
Step 1: Calculate the pressure at the depth
The hydrostatic pressure [tex]\( P_{\text{water}} \)[/tex] at a depth h below the ocean surface is given by:
[tex]\[P_{\text{water}} = \rho g h\][/tex]
Calculate [tex]\( P_{\text{water}} \):[/tex]
[tex]\[P_{\text{water}} = 1025 \times 9.81 \times 28\][/tex]
[tex]\[P_{\text{water}} = 285396 \, \text{Pa}\][/tex]
Step 2: Calculate the total pressure inside the submersible
The total pressure inside the submersible is the sum of the atmospheric pressure and the hydrostatic pressure at depth:
[tex]\[P_{\text{total}} = P_{\text{inside}} + P_{\text{water}}\][/tex]
[tex]\[P_{\text{total}} = 1.0 \times 1.013 \times 10^5 + 285396\][/tex]
[tex]\[P_{\text{total}} = 1.013 \times 10^5 + 285396\][/tex]
[tex]\[P_{\text{total}} = 386396 \, \text{Pa}\][/tex]
Step 3: Calculate the force required to open the hatch
The force F required to open the hatch against the pressure difference is:
[tex]\[F = (P_{\text{total}} - P_{\text{atm}}) \times A\][/tex]
Substitute the values:
[tex]\[F = (386396 - 1.013 \times 10^5) \times 0.70\][/tex]
[tex]\[F = (386396 - 101300) \times 0.70\][/tex]
[tex]\[F = 285096 \times 0.70\][/tex]
[tex]\[F = 199567.2 \, \text{N}\][/tex]
The position of a particle moving along the x axis depends on the time according to the equation x = ct3 - bt6, where x is in meters and t in seconds. Let c and b have numerical values 2.9 m/s3 and 2.0 m/s6, respectively. From t = 0.0 s to t = 1.3 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.
Answer:
Given the equation of the particle, we know that:
[tex]x=(2.9\frac{m}{s^{3} } )*t^{3} -(2\frac{m}{s^{6} } )*t^{6}[/tex]
a) In [tex]t=0.0s[/tex] we evaluate [tex]t[/tex] in the former equation:
[tex]x_{1} =(2.9\frac{m}{s^{3} } )*(0)^{3} -(2\frac{m}{s^{6} } )*(0)^{6}=0m[/tex]
In [tex]t=1.3s[/tex]
[tex]x_{2} =(2.9\frac{m}{s^{3} } )*(1.3s)^{3} -(2\frac{m}{s^{6} } )*(1.3s)^{6}=-3,28232m[/tex]
a) So, we know that the displacement of te particle is given by:
[tex]x=x_{2}-x_{1} =-3.28232m-0m=-3.28232m[/tex]
To find it's velocity, we need to derivate the equation of position by the formula:
[tex]v_{x} =\frac{dx}{dt} =3ct^{2}-6bt^{5} =(8.7\frac{m}{s^{3} })t^{2} -(12\frac{m}{s^{6} } )t^{5}[/tex]
And evaluate this expression at each specified t:
b) [tex]v(1s)_{x} =(8.7\frac{m}{s^{3} })(1s)^{2} -(12\frac{m}{s^{6} } )(1s)^{5}=-3.3\frac{m}{s}[/tex]
c) [tex]v(2s)_{x} =(8.7\frac{m}{s^{3} })(2s)^{2} -(12\frac{m}{s^{6} } )(2s)^{5}=-349.2\frac{m}{s}[/tex]
d) [tex]v(3s)_{x} =(8.7\frac{m}{s^{3} })(3s)^{2} -(12\frac{m}{s^{6} } )(3s)^{5}=-2837.7\frac{m}{s}[/tex]
e) [tex]v(4s)_{x} =(8.7\frac{m}{s^{3} })(4s)^{2} -(12\frac{m}{s^{6} } )(4s)^{5}=-12148.8\frac{m}{s}[/tex]
To find it's acceleration, we need to derivate the equation of velocity by the formula:
[tex]a_{x} =\frac{dv}{dt} =(17.4\frac{m}{s^{3} } )t-(60\frac{m}{s^{6} } )t^{4}[/tex]
And evaluate this expression at each specified t:
f) [tex]a(1s)_{x} =(17.4\frac{m}{s^{3} } )(1s)-(60\frac{m}{s^{6} } )(1s)^{4}=-42.6\frac{m}{s^{2} }[/tex]
g) [tex]a(2s)_{x} =(17.4\frac{m}{s^{3} } )(2s)-(60\frac{m}{s^{6} } )(2s)^{4}=-925.2\frac{m}{s^{2} }[/tex]
h) [tex]a(3s)_{x} =(17.4\frac{m}{s^{3} } )(3s)-(60\frac{m}{s^{6} } )(3s)^{4}=-4807.8\frac{m}{s^{2} }[/tex]
i) [tex]a(4s)_{x} =(17.4\frac{m}{s^{3} } )(4s)-(60\frac{m}{s^{6} } )(4s)^{4}=-15290.4\frac{m}{s^{2} }[/tex]
Determine whether each of the following electron configurations is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choices.(a) 1s22s22p63s23p5(b) 1s22s22p63s23p63d74s2(c) 1s22s22p63s23p63d104s24p6(d) 1s22s22p63s23p64s1(e) 1s22s22p63s23p63d104s24p64d55s2(f) 1s22s22p63s2
The electron configurations correspond to a halogen, transition metals, an inert gas, an alkali metal, and an alkaline earth metal, identified by their completed electron orbitals and the presence of electrons in specific orbitals such as d or p.
Explanation:To determine whether an electron configuration corresponds to an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal, you must look at the position that the electron configuration would occupy on the periodic table.(a) 1s2 2s2 2p6 3s2 3p5: This is a halogen because it is one electron short of a noble gas configuration, indicating it is in group 17.(b) 1s2 2s2 2p6 3s2 3p6 3d7 4s2: This is a transition metal due to the partially filled d orbital.(c) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6: This is an inert gas because all orbitals are filled, indicating that it is in group 18.(d) 1s2 2s2 2p6 3s2 3p6 4s1: This is an alkali metal, in group 1, with a single electron in the outermost s orbital.(e) 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d5 5s2: This configuration belongs to a transition metal because of the partially filled d orbital (4d5).(f) 1s2 2s2 2p6 3s2: This is an alkaline earth metal with two electrons in the outermost s orbital, indicating it is in group 2.A photographer in a helicopter ascending vertically at a constant rate of accidentally drops a camera out the window when the helicopter is 60.0 m above the ground. (a) How long will the camera take to reach the ground?
Answer:
The time it takes for the camera to reach the ground is 5 s.
Explanation:
To solve this problem, we will use the free fall cinematic equation.
Since the helicopter ascends with constant speed, the camera falls to the ground only by the effect of gravity on it.
The speed at which the helicopter ascends is not specified in the statement, but according to a similar problem, we will use 12.5 [tex]\frac{m}{s}[/tex].
First, we must calculate the time and the maximum height at which the camera arrives after leaving the helicopter.
To calculate the maximum height to which it arrives, we will use the formula of vertical shot (since the camera leaves the helicopter with a speed upwards of 12,5 [tex]\frac{m}{s}[/tex]).
[tex]V_{f} ^{2}[/tex]=[tex]V_{0} ^{2}[/tex] - 2 * g * h
Where:
[tex]V_{f}[/tex]: final speed at maximum height.
[tex]V_{0}[/tex]: initial speed when it falls from the helicopter.
g: gravity taken at 9.8 [tex]\frac{m}{s^{2} }[/tex]
h: height reached from 60 m when leaving the helicopter
as [tex]V_{f}[/tex]=0
0=[tex](12,5\frac{m}{s}) ^{2}[/tex] - 2 * [tex]9,8\frac{m}{s^{2} }[/tex] * h
clear h:
h=[tex](12,5 \frac{m}{s^{2} } )^{2}[/tex] / (2 * [tex]9,8 \frac{m}{s^{2} }[/tex])
h=7,97 m
Then we must calculate the time it takes to reach its maximum height:
[tex]V_{f}[/tex]=[tex]V_{0}[/tex] - g * t
t: time it takes to arrive from the moment it leaves the helicopter at its maximum height.
as [tex]V_{f}[/tex]=0
0=12.5 [tex]\frac{m}{s}[/tex] - 9.8 [tex]\frac{m}{s^{2} }[/tex] * t
clearing t
t=12.5 [tex]\frac{m}{s}[/tex] / 9.8 [tex]\frac{m}{s^{2} }[/tex]
t=1.27 s.
Now we can calculate the time it takes to fall from the maximum height of 67.97 m.
The equation we will use is Y=[tex]v_{0}*t+(\frac{g*t^{2} }{2} )[/tex]
where:
t: time it takes for the camera to fall.
Y: height from where the camera falls concerning the ground.
[tex]v_{0}[/tex]: initial speed of the camera at the time of starting the fall.
g: acceleration of gravity, estimated at 9.8 [tex]\frac{m}{s^{2} }[/tex]
Step 1: As the helicopter ascends with constant speed, the initial speed of the camera at the moment of falling is 0.
[tex]v_{0}[/tex]=0
So the first term of our equation is nullified.
Step 2: To calculate the time it takes to fall, we clear "t" of the equation:
Y=[tex]\frac{(g*t^{2})}{2}[/tex]
Y*2=(g*[tex]t^{2}[/tex])
[tex]\frac{Y*2}{g}[/tex]=[tex]t^{2}[/tex]
[tex]\sqrt{\frac{Y*2}{g} }[/tex]=t
Step 3: I replace the values with the incognites and get "t".
t=[tex]\sqrt{\frac{67,97m*2}{9,8\frac{m}{s^{2} } } }[/tex]
t=3,73 s
The total time it takes for the camera to fall from the moment it leaves the helicopter is the sum of the time it takes to reach the maximum point of height and the time it takes to fall to the ground from that height.
t= 1,27 s + 3,73 s = 5 s
Have a nice day!
Final answer:
The camera will take approximately 3.9 seconds to reach the ground.
Explanation:
To find the time it takes for the camera to reach the ground, we can use the equation of motion for an object in free fall:
h = (1/2)gt²
where h is the initial height (60.0 m), g is the acceleration due to gravity (9.8 m/s²), and t is the time. Since the camera is dropped, its initial velocity is zero. Rearranging the equation, we have:
t = [tex]\sqrt{(2h / g)[/tex]
Substituting the given values:
t = [tex]\sqrt{2 * 60.0 / 9.8)[/tex]
t ≈ 3.9 seconds
So, the camera will take approximately 3.9 seconds to reach the ground.
A solid sample of a metal is submerged in a large graduated cylinder. The water level in the graduate was 55.25 mL and the final level was 61.00 mL. If the mass of the object was determined to be 23.795 g, what is the density of the metal?
Answer:
ρm=4.14g/ml :metal density
Explanation:
Conceptual analysis
We apply the formula to calculate the density:
ρ= m /V Formula (1)
Where:
ρ:density in g/mL
m: mass in g (grams)
V= volume in ml (milliliters)
Known data
[tex]V_{il}[/tex]: final volume of liquid=55.25 ml
[tex]V_{fl}[/tex]: final volume of liquid=61.00 ml
ms: metal mass=23.795 g
Problem development
The volume of the metal (vm) is equal to the change in volume of the liquid.
[tex]V_{m} = V_{fl}-V_{il} =61.00 ml - 55.25 ml =5.75ml[/tex]
We replace the data in formula (1 )to calculate the density of the metal(ρm):
ρm= mm /Vm=23.795 g/5.75ml
ρm=4.14g/ml
You observe a light ray move from one piece of glass to another (a different type of glass) and the light ray bends towards the glass-glass interface (away from the normal) when it enters the 2nd piece of glass. You deduce that:
Answer:
second medium is less optical denser than first medium
Explanation:
As per Snell's law we know that
[tex]\mu_1 sin\theta_1 = \mu_2 sin\theta_2[/tex]
so here we know that
[tex]\theta_1 [/tex] = angle of incidence
[tex]\theta_2[/tex] = angle of refraction
As we know that light ray goes away from the normal when it move from medium 1 to medium 2
so here angle of refraction will me more than the angle of incidence
so we have
[tex]\frac{sin\theta_1}{sin\theta_2} = \frac{\mu_2}{\mu_1}[/tex]
since we know that
[tex]\theta_1 < \theta_2[/tex]
so we have
[tex]\frac{sin\theta_1}{sin\theta_2} < 1[/tex]
[tex]\frac{\mu_2}{\mu_1} <1[/tex]
so we can say that
[tex]\mu_2 < \mu_1[/tex]
so we can say that second medium is less optical denser than first medium
A straight, nonconducting plastic wire 9.50 cm long carries a charge density of 130 nC/m distributed uniformly along its length. It is lying on a horizontal tabletop.A) Find the magnitude and direction of the electric field this wire produces at a point 4.50 cm directly above its midpoint.B)If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 4.50 cm directly above its center.
Answer:
A) E = 3.70*10^{4} N/C
B) E = 2.281*10^3 N/C
Explanation:
given data:
charge density [tex] \lambda = 130*10^{-9} C/m[/tex]
length of wire = 9.50 cm
a) at x = 4.5 m above midpoint, electric field is calculated as
[tex]E = \frac{1}{ 2\pi \epsilon} * \frac{ \lambda}{x\sqrt{(x^2/a^2)+1}}[/tex]
x = 4.5 cm
midpoint a = 4.5 cm = 0.0475 m
[tex]E =2{\frac{1}{ 8.99*10^9} * \frac{130*10^{-9} }{0.045\sqrt{(4.5^2/4.75^2)+1}}[/tex]
E = 3.70*10^{4} N/C
B) when wire is in circle form
[tex]Q = \lambda * L[/tex]
[tex]= 130*10^{-9} *9.5*10^{-2}[/tex]
= 1.235*10^{-8} C
Radius of circle
[tex]r = \frac{L}{2\pi}[/tex]
[tex]r = \frac{9.5*10^{-2}}{2\pi}[/tex]
r = 1.511*10^{-2} m
[tex]E = \frac{1}{ 2\pi \epsilon} * \frac{Qx}{(x^2+r^2)^{3/2}}[/tex]
[tex]E =8.99*10^{9} * \frac{1.23*10^{-8}*4.5*10^{-2}}{((4.5*10^{-2})^2+(1.511*10^{-2})^2)^{3/2}}[/tex]
E = 2.281*10^3 N/C
Think of an example that you could use to convice a friend that an object at rest has an "internal hold back property" that is different from a frictional force or the force of the earth's gravity.
Answer:
Explanation:
There are various examples around us that states " internal hold back property " and is different from frictional force or the force of Earth's gravity.
The plaster which is the mixture of cement and water has an adhesive nature and molecules inside them has a internal hold back property. They stick to the walls and ceiling and do not fall.
Though gravity is applied on it still its molecules tends to hold the ceiling firmly and do not fall upon.
Technician A says that a diesel engine’s higher compression ratio helps it to produce more power and use less fuel. Technician B says that higher compression ratios only help the diesel engine produce more torque. Who is correct?
Answer:
None of them is 100% correct.
Technician A is correct when he says that the higher compression rate helps the car to use less fuel. A higher compression ratio gives the engine a higher thermal efficiency which in turn translates to higher fuel efficiency but it does not produce more power than a gasoline engine.
Technician B is correct when he says the diesel engine produces more torque, but he also says that it is the only thing that is better with the diesel engine which is wrong because we already know that it also helps on fuel efficiency.
Two of the types of infrared light, IR-C and IR-A, are both components of sunlight. Their wavelengths range from 3000 to 1,000,000 nm for IR-C and from 700 to 1400 nm for IR-A. Compare the energy of microwaves, IR-C, and IR-A.
Explanation:
3000 to 1,000,000 nm for IR-C
700 to 1400 nm for IR-A
Now, we are working with light, so we can use that the energy of a photon with frequency f is : E(f) = h*f
where h is the plank constant h = 6.62607015×10−34 j*s
and f is the frequency that f= c/λ, where λ is the wavelenght.
so, for IR-C the energy lies between h*c/3000 and h*c/1000000
where c is the light speed.
For IR-A the energy lies between h*c/700 and h*c/1400.
so here you can se that IR-A has a lot more energy than IR-C, you can se that the minimal energy of IR-A is h*c/1400 and the maximal of IR-C is h*c/3000 so the minimal energy of IR-A is almost twice times the maximum energy of IR-C
On earth, a block is placed on a frictionless table on earth. When a horizontal force of 10 N is applied to the block, it accelerates at 5.3 m/s2. Suppose the block and table are set up on the moon. When a horizontal force of 5 N is applied to the block, what is the acceleration?
a.3.2m/s^2
b. 2.7m/s^2
c. 3.4m/s^2
d. 2.4m/s^2
Answer:
The answer to your question is: b. 2.7m/s²
Explanation:
Data
Force = 10 N
a = 5.3 m/s²
Moon
F = 5 N
a = ?
Formula
F = m x a
Process
Find the mass of the table
m = F / a
m = 10 / 5.3
m = 1.887 kg
Now, find the acceleration
a = F / m
a = 5 / 1.887
a = 2.65 m/s² ≈ 2.7 m/s²
Answer:
a = F / m = 5N / 1.887kg = 2.65 m/s (answer choice B)
Explanation:
rounded the answer
An American manufacturer supplied a customer with refrigerators with electrical cords that were one yard long instead of 1 meter long. What was the difference in cord length (cm)?a. 0.856 cmb. 85.6 cmc. 8.56 cmd. 0.00856 cme. 856 cm
Answer:
C.
Explanation:
A meter is 8.56 centimeters longer than a yard. Something to keep in mind is that a meter is about 10% longer than a yard.
Hope this helps :)
Answer:
C. 8.56cm
Explanation:
Since we are to get the difference of cord length in centimeter, there fore we will have to convert the one yard and one metre length of wires to centimeters.
For 1m length of cord
1m = 100cm
For 1yard length of cord
1yard = 91.44cm
Difference in cord length = 100cm - 91.44cm
= 8.56cm
What is the entropy change of a 29.8 g ice cube that melts completely in a bucket of water whose temperature is just above the freezing point of water?
Answer:
[tex]\Delta S = 36.55 J/K[/tex]
Explanation:
As we know that entropy change for phase conversion is given as
[tex]\Delta S = \frac{\Delta Q}{T}[/tex]
Here we know that heat required to change the phase of the ice is given as
[tex]\Delta Q = mL[/tex]
here we have
m = 29.8 g = 0.0298 kg
L = 335000 J/kg
now we have
[tex]\Delta Q = 0.0298\times 335000[/tex]
[tex]\Delta Q = 9979.4 J[/tex]
also we know that temperature is approximately same as freezing temperature
so we have
[tex]T = 273 k[/tex]
so here we have
[tex]\Delta S = \frac{9979.4}{273}[/tex]
[tex]\Delta S = 36.55 J/K[/tex]
two mechanical devices typically used in laboratories to accurately measure small objects or distances are the _____ and _____. a)microscope b)meter stick c)micrometerd) calipers
Answer:
Option c and d
Explanation:
The two mechanical devices which are used in laboratories for accurate measurement of small distances or small objects are calipers and micrometer.
Micrometer is a mechanical device used in laboratories like that of a screw gauge. It is used for the measurement of thickness of objects, length and the depth of the small objects which can be measured by holding the object in between the spindle and anvil of the micrometer.
Calipers is another mechanical device like that of vernier calipers used in laboratories for measurement of small distances, usually the distance between the opposing faces of the object. The measurement is usually taken on a digital display, a dial or a scale that is ruled.
The distance is measured by adjusting the tips of the caliper holding the object on a ruler.
A student driving home for the holidays starts at 8:00 am to make the 675-km trip, practically all of which is on nonurban interstate highway. If she wants to arrive home no later than 3:00 pm, what must be her minimum average speed? Will she have to exceed the speed limit?
8 am to 3 pm is 7 hours.
(675 km) / (7 hrs) = 96.4 km/hr .
Her average speed for the whole 7 hours has to be not less than 96.4 km/hr. Any less, she's not home by 3:00.
We don't technically know the speed limit at every point on her trip, because you technically haven't told us.
But in 7 hours, she MUST stop for gas, she MUST get some rest, she MUST make a pit stop, and she most likely encounters some traffic somewhere. So in order to average 96.4 for the whole trip, she MUST exceed it for PART of the trip ... possibly by a lot.
Whatever the speed limit may be, I think it's likely that she'll exceed it SOMEwhere, at least for SOME time.
The student must maintain a minimum average speed of 96.43 km/h.
Whether this exceeds the speed limit depends on the highway's speed limit.
To determine the minimum average speed the student needs to maintain to complete the 675-km trip by 3:00 pm, starting at 8:00 am,
We first calculate the total travel time allowed, the difference between 3:00 pm and 8:00 am is 7 hours.
Therefore, to find the minimum average speed, we divide the total distance by the total time:
Average Speed = Total Distance / Total Time
Average Speed = 675 km / 7 hours
= 96.43 km/h
The student must maintain a minimum average speed of 96.43 km/h to reach the destination on time.
A light-year is the distance that light travels in one year. The speed of light is 3.00 × 108 m/s. How many miles are there in one light-year? (1 mi = 1609 m, 1 y = 365 d)
a 5.88 × 1012 mi
b 9.46 × 1015 mi
c 5.88 × 1015 mi
d 9.46 × 1012 mi
Answer:
(a) [tex]5.88\times 10^{12}mi[/tex]
Explanation:
We have given speed of light [tex]c=3\times 10^8m/sec[/tex]
Given time = 1 year =365 days
We know that in 1 minute = 60 sec
1 hour = 60×60 = 3600 sec
In one day = 24 hour = 24×60×60=86400 sec
So in 365 days = 365×86400[tex]=3.1536\times 10^7sec[/tex]
We know that distance = speed ×time[tex]=3.1536\times 10^7\times 3\times 10^8=9.46\times 10^{15}m[/tex]
We have given that 1 mi = 1609 m
So [tex]9.46\times 10^{15}m=\frac{9.46\times 10^{15}}{1609}=5.88\times 10^{12}mi[/tex]So option (a) is correct