A quarterback back pedals 3.3 meters southward and then runs 5.7 meters northward. For this motion, what is the distance moved? What is the magnitude and direction of the displacement?

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

[tex]V=X/T[/tex]

First stage:

T1=5s

[tex]v_{f}  =v_{o} - at[/tex]

But, [tex]v_{f}  =0[/tex]   (decelerates to rest)

then: [tex]a =v_{o} /t=0.3/5=0.06m/s^{2}[/tex]

on the other hand:

[tex]x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m[/tex]

X1=75cm

Second stage:

T2=5s

[tex]x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m[/tex]

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

The average speed of the particle is 20 centimetre per second.

Average speed is given as total distance covered by the particle divided by the total time of the journey.

The particle decelerates from 30 cm/s to rest (0 cm/s) in 5.0 s. Hence, the deceleration can be given as,

a = (0 - 30 cm/s) / 5 s = - 6 cm/s²

From the kinematic equation:

s = ut + [tex]\frac{1}{2}at^2[/tex], where s = displacement, a = acceleration, t = time, and u = initial velocity of the particle, we can determine the displacement.

so, s = (30 cm/s) (5 s) - [tex]\frac{1}{2} (6 cm/s^2)(5 s)^2[/tex] = 150 cm - 75 cm = 75 cm

The particle then accelerates uniformly at 10 cm/s² for another 5.0 s. The final velocity after this period is given as:

v = u + at

Here, u = 0 cm/s, a = 10 cm/s². t = 5s. Hence,

v = (10 × 5) m/s = 50 cm/s

Using the kinematic equation:

v² = u² + 2as, we get:

(50 cm/s)² = 2 (10 cm/s²) s

or, s = 125 cm

Total displacement of the particle = 75 cm + 125 cm = 200 cm

total time of journey = 5s + 5s = 10s

so, average velocity = 200 cm / 10 s = 20 cm/s

Answer:

New force, F' = F

Explanation:

Given that, two small balls, A and B, attract each other gravitationally with a force of magnitude F. It is given by :

[tex]F=G\dfrac{m_Am_B}{r^2}[/tex]

If we now double both masses and the separation of the balls, the new force is given by :

[tex]F'=G\dfrac{2m_A\times 2m_B}{(2r)^2}[/tex]

F' = F

So, the new force remains the same as previous one. Hence, the correct option is (d) "F"

Answer:

After studying the law of gravitational attraction, students constructed a model to illustrate the relationship between gravitational attraction (F) and distance. If the distance between two objects of equal mass is increased by 2, then the gravitational attraction (F) is 1/4F or F/4. How would this model, situation A, change if the mass of the spheres is doubled?

A)  A

B)  B

C)  C

D)  D

If you came here from usa test prep it is:

Actually A

But for the question given right now is D.

Explanation:

Answer:

customer loyalty

Explanation:

According to my research on five forces model, I can say that based on the information provided within the question the two strength factors that relate to all three of the competitive forces are switching costs and​ customer loyalty. Customer loyalty is when a customers choose and become loyal to a certain business over their competitors because their organizational culture and customer service.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

Answer:

A. Scatterplot

Explanation:

because it is

Answer: Technician B

Explanation: A taillights, and several kind of ilumation devices, works by converting a current flow trought a material in Light. This mean that the Current Flow (Amperes) which goes through a lights is proportional to the light it gives (Lumens).

This means, that if the tailights has a dimly light the current that goes through is less than the one it is suppose to be.

A fuse is a protection device for over-current, if the current at any given time goes beyond a limit (designed on the fuse) the fuse will melt and cutting all the current to the circuit. A blown fuse will cut all the current from the circuit and the tailights will be completely off.

However a open ground circuit is a differente kind of failure. In this cases, there is a pact from where the current on the circuit "escapes" from it. This could be by several reason, unprotected wires the most usual. While the current escapes from his intended course, not all of the energy goes away from the load. This explain why the tailight still has enough energy through it to light dimly.

The estimated diameter of the Moon is approximately [tex]3.536 \times 10^4[/tex] meters, (two significant figures as [tex]3.5 \times 10^4[/tex] meters).

Here, we have to set up a proportion to solve for the diameter of the Moon using the given information:

The pencil's diameter = 0.7 cm

Distance from the eye to the pencil = 0.75 m

Earth-Moon distance = [tex]3.8 \times 10^5[/tex] km = [tex]3.8 \times 10^8[/tex] m

Let, the diameter of the Moon as "D."

When the pencil blocks out the Moon, the ratio of the pencil's diameter to the distance from the eye to the pencil is equal to the ratio of the Moon's diameter to the Earth-Moon distance:

Diameter of Moon / Earth-Moon distance = Pencil's diameter / Distance to pencil

D / (  [tex]3.8 \times 10^8[/tex] m) = (0.7 cm) / (0.75 m)

Now, solve for D:

D = (0.7 cm) * ( [tex]3.8 \times 10^8[/tex] m) / (0.75 m)

D ≈ [tex]3.536 \times 10^6[/tex] cm

Convert the diameter of the Moon to meters:

D ≈ [tex]3.536 \times 10^4[/tex] m

So, the estimated diameter of the Moon is approximately [tex]3.536 \times 10^4[/tex] meters, which can be expressed using two significant figures as [tex]3.5 \times 10^4[/tex] meters.

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Answer:

the potential energy increases and the electric potential increases

Explanation:

we know here that Voltage i.e. (Electric potential) increases from the initial point  to the final point

consider if 2 object of different charge 1 is twice the charge of other moveing in same distance in electric filed

than object of twice the charge require twice the force so twice amount of work

This work change the potential energy by  equal amount of work done

so electric potential energy is depend on the amount of charge on object

so if work being done on a positive test charge by an external force in moving the charge from one location to another

the potential energy increases and the electric potential increases

When a positive test charge is moved from one location to another, the potential energy increases and the electric potential energy increases.

The work done in moving a unit positive test charge from infinity to a certain point in the electric field is known as potential energy.

V = E x d

V = (F/q) x d

where;

Thus, when a positive test charge is moved from one location to another, the potential energy increases and the electric potential energy increases.

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Answer:

The natural, fundamental frequency of this string is 24 hertz.          

Explanation:

Length of the string, l = 2 m

Number of segments, n = 5

Frequency, f = 120 Hz

Let f' is the natural fundamental frequency of this string. The frequency for both side ended string is given by :

[tex]f=\dfrac{nv}{2l}[/tex]

[tex]v=\dfrac{2fl}{n}[/tex]

[tex]v=\dfrac{2\times 120\times 2}{5}[/tex]        

v = 96 m/s

For fundamental frequency, n = 1

[tex]f'=\dfrac{v}{2l}[/tex]

[tex]f'=\dfrac{96}{2\times 2}[/tex]

f' = 24 Hertz

So, the natural, fundamental frequency of this string is 24 hertz. Hence, this is the required solution.

The natural, fundamental frequency of this standing wave is equal to 24 Hertz.

Given the following data:

Length of the string = 2.0 meters.

Number of segments = 5.

Frequency = 120 Hertz.

First of all, we would determine the velocity of the standing wave by using this formula:

[tex]F=\frac{nV}{2L} \\\\V=\frac{2FL}{n} \\\\V=\frac{2 \times 120 \times 2.0}{5}\\\\V=\frac{480}{5}[/tex]

V = 96 m/s.

Now, we can calculate the natural, fundamental frequency by using this formula:

[tex]F'=\frac{V}{2L} \\\\F'=\frac{96}{2 \times 2.0} \\\\F'=\frac{96}{4.0}[/tex]

F' = 24 Hertz.

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Answer:

[tex]\theta=41.52^{\circ}[/tex]

Explanation:

Given that,

Velocity of the  airplane, v = 240 m/s

Angle with horizontal, [tex]\theta=30^{\circ}[/tex]

The altitude of the plane is 2.4 km, d = 2400 m

Vertical speed of the airplane, [tex]v_y=v\ sin\theta=240\ sin(30)=120\ m/s[/tex]

Horizontal speed of the airplane, [tex]v_x=v\ cos\theta=240\ sin(30)=207.84\ m/s[/tex]

So, the equation of the projectile for the flare is given by :

[tex]4.9t^2+120t-2400=0[/tex]

On solving the above equation, we get the value of t as:

t = 13.04 seconds

Horizontal distance travelled,

[tex]d=v_x\times t[/tex]

[tex]d=207.84\times 13.04[/tex]

d = 2710.23 m

Let [tex]\theta[/tex] is the angle with which it hits the target. So,

[tex]tan\theta=\dfrac{2400}{2710.23}[/tex]

[tex]\theta=41.52^{\circ}[/tex]

Hence, this is the required solution.

To solve this problem, you would apply principles of physics like projectile motion and trigonometry. We calculate the horizontal and vertical velocities using the given initial velocity and angle. The final total velocity and angle can be found by using these calculations as the horizontal velocity does not change.

The subject requires the application of the concepts of physics, specifically kinematics and trigonometry. Understanding the question in context, we are given that the airplane is flying at a velocity of 240m/s at an angle of 30.0° with the horizontal. A flare is released from the plane when it is at an altitude of 2.4km, and it hits a target on the ground. The problem needs us to find the angle θ.

Considering the fact that the time for projectile motion is completely determined by vertical motion, we set up the problem in the following way: We break down the initial velocity into components using the initial angle. The horizontal velocity (Vx) can be calculated using Vx = V*cos(θ), and the vertical velocity (Vy) can be calculated using Vy = V*sin(θ), where V is the initial velocity and θ is the initial angle.

Since the horizontal motion is constant and the initial position is known, we can use these two vertical and horizontal velocities to find the total velocity and the angle it makes with the horizontal. The trick here is to remember that since the x component (horizontal velocity) doesn't change, we can determine the final total velocity and its angle using these components.

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Answer:

These are the answers for 1, 2 and 3

Explanation:

Sorry I couldn't help you with 4 and 5

1.

Answer:

y = 11.48 m

x = 13.0 m

Explanation:

Components of initial velocity is given as

[tex]v_x = 7.2 cos25 = 6.52 m/s[/tex]

[tex]v_y = 7.2 sin25 = 3.04 m/s[/tex]

Now after t = 2 s the vertical position is given as

[tex]y = y_0 + v_y t + \frac{1}{2}a_y t^2[/tex]

[tex]y = 25 + 3.04(2) - \frac{1}{2}(9.8)(2^2)[/tex]

[tex]y = 11.48 m[/tex]

Now horizontal position is given as

[tex]x = v_x t[/tex]

[tex]x = 6.52 \times 2[/tex]

[tex]x = 13.04 m[/tex]

2.

Answer:

d = 26.6 m

Explanation:

Initial position on y axis is given as

[tex]y = 1.5 m[/tex]

velocity of ball in y direction

[tex]v_y = 0[/tex]

now we have

[tex]\Delta y = v_y t + \frac{1}{2}gt^2[/tex]

[tex]1.5 = \frac{1}{2}(9.8) t^2[/tex]

[tex]t = 0.55 s[/tex]

now the distance moved by the ball in horizontal direction is given as

[tex]d = v_x t[/tex]

[tex]d = 48.1 \times 0.55[/tex]

[tex]d = 26.6 m[/tex]

3

Answer:

[tex]A + B = 12.42\hat i + 0.35 \hat j[/tex]

Explanation:

Here we can see the two vectors inclined at different angles

so two components of vector A is given as

[tex]A = 11.3 cos21\hat i - 11.3 sin21\hat j[/tex]

[tex]A = 10.55 \hat i - 4.05 \hat j[/tex]

Similarly for other vector B we have

[tex]B = 4.78cos67 \hat i + 4.78 sin67\hat j[/tex]

[tex]B = 1.87\hat i + 4.4 \hat j[/tex]

now we need to find A + B

so we have

[tex]A + B = (10.55\hat i - 4.05\hat j) + (1.87\hat i + 4.4 \hat j)[/tex]

[tex]A + B = 12.42\hat i + 0.35 \hat j[/tex]

4.

Answer:

d = 81.86 m

Explanation:

Displacement of airplane is given as

[tex]d_1[/tex] = 72 km in direction 30 degree East of North

[tex]d_1 = 72sin30\hat i + 72cos30\hat j[/tex]

[tex]d_1 = 36\hat i + 62.35\hat j[/tex]

[tex]d_2[/tex] = 48 km South

[tex]d_2 = -48\hat j[/tex]

[tex]d_3[/tex] = 100 km in direction 30 degree North of West

[tex]d_3 = -100 cos30\hat i + 100 sin30\hat j[/tex]

[tex]d_3 = -86.6\hat i + 50\hat j[/tex]

so net displacement is given as

[tex]d = d_1 + d_2 + d_3[/tex]

[tex]d = 36\hat i + 62.35\hat j - 48\hat j - 86.6\hat i + 50 \hat j[/tex]

[tex]d = -50.6\hat i + 64.35\hat j[/tex]

now magnitude of displacement is given as

[tex]d = \sqrt{50.6^2 + 64.35^2}[/tex]

[tex]d = 81.86 m[/tex]

5.

Answer:

1) final speed at which it will hit the ground again

2) acceleration during the motion of the ball

Explanation:

As we know that the speed at which the ball is thrown is always same to the speed by which it will hit back on the ground

so we know that final speed will be same as initial speed

Also we know that during the motion the acceleration of ball is due to gravity so it will be

[tex]a = - g[/tex]

Answer:

- 3.33 m /s

Explanation:

m1 = 5 kg

m2 = 10 kg

u1 = 10 m/s

u2 = - 10 m/s

Let the velocity of the combination of the blob is v.

Use the conservation of momentum

[tex]m_{1}u_{1}+m_{2}u_{2}=\left ( m_{1}+m_{2} \right )v[/tex]

5 x 10 - 10 x 10 = (5 + 10)v

50 - 100 = 15 v

v = - 3.33 m /s

Thus, the velocity of the blob after sticking together is - 3.33 m/s.

Answer:

The required radius of its motion is [tex]0.084m[/tex].

Explanation:

The formula for calculating the required  radius of its motion is given by

[tex]F = (mv^2)/r[/tex]

Where m= mass  

V= moving velocity

F=frictional force

r = radius of its motion

Then the required radius of its motion is given by

[tex]r =(mv^2)/F[/tex]

Given that

mas =0.0818 kg

Frictional force= 0.108 N

Moving with Velocity of  = 0.333 m/s

radius of its motion = [tex]\frac{[0.0818 kg \times (0.333 m/s)^2]}{0.108 N}[/tex]

Hence the required radius of its motion is r = [tex]8.4 cm=0.084m[/tex]

Answer:

0.084

Explanation:

Acellus

Answer:

35 meters

Explanation:

The river flows at 2 m/s from North to South and motor can propel the boat at 5m/s. As the downstream direction is positive, we are considering the river flow also propels the boat adding its speed to the boat. It means the boat and the person in the inner tube are in fact moving at 7 m/s. Distance can be calculated as follows:

[tex]v = d/t[/tex]  ⇒[tex]d = vt[/tex] ⇒[tex]d = 7\frac{m}{s}x5s[/tex]

[tex]d = 35m[/tex]

The motorboat will cover a distance of 35 meters downstream in 5 seconds. This is calculated by adding the boat's velocity relative to the water (5 m/s) to the velocity of the river current (2 m/s), giving a resultant velocity of 7 m/s and multiplying it by the time interval.

We need to calculate the distance the motorboat will cover downstream in a 5-second interval. The speed of the river current is 2 m/s and the speed of the motorboat relative to the water is 5 m/s. Since we are considering the downstream direction as positive, the boat's velocity relative to an observer on the shore would be the sum of these two speeds.

The boat's speed relative to the shore (resultant velocity) is:

Velocity of the boat relative to the shore = Velocity of the boat relative to the water + Velocity of the river = 5 m/s + 2 m/s = 7 m/s

The distance covered by the boat over a period of 5 seconds would be:
Distance = Velocity × Time = 7 m/s × 5 s = 35 m

Therefore, the motorboat will move 35 meters downstream as observed by someone outside of the water, like the person in the inner tube.

Answer:

a) [tex]V=14.01 m/s[/tex]

b) [tex]V=8.86 \, m/s[/tex]

c)[tex]t = 2.33s[/tex]

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

[tex]E_m = E_k +E_p[/tex]

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this [tex]E_m[/tex] mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

[tex]E_k = \frac{1}{2} m \, V^2[/tex]

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

[tex]E_p= m\, g\, h[/tex]

where g is the acceleration due to gravity ([tex]g=9.81 \, m/s^2[/tex]) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which [tex]h=0[/tex], a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

[tex]E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m[/tex]

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set [tex]h=0[/tex]

[tex]E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2[/tex]

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

[tex]E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\[/tex]

And so we have found the velocity of the ball as it hits the floor.

[tex]V = \sqrt[]{2g\cdot 10m}=14.01\, m/s[/tex]

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

[tex]V = \sqrt[]{2g\cdot h}[/tex]

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

[tex]V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s[/tex]

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

[tex]V_{avg}\cdot t=h\\t=h/V_{avg}[/tex]

what's the average speed when the ball is descending?

[tex]V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s[/tex]

so the time it takes the ball to go down is:

[tex]t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\[/tex]

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

[tex]V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s[/tex]

and the time it takes to go up is:[tex]t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s[/tex]

When we add both times , we get:

[tex]t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s[/tex]

To find the speed of the tennis ball just before it hits the floor on its way down, we can use the equation y = 0 + voyt - 1/2gt^2. The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec and end at v = -0.98 m/s at t = 0.65 sec. The time the ball is in the air from the time it is dropped until it reaches its maximum height on the first rebound is 2.5 s.

(a) While the ball is in the air, it rises and then falls to a final position 10.0 m higher than its starting altitude. We can find the time for this by using Equation 4.22: y = yo + voyt - 1/2gt^2. If we take the initial position yo to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

(b) The graph of the ball's vertical velocity over time should begin at 4.90 m/s during the time interval 0 - 0.1 sec (there should be a data point at t = 0.05 sec, v = 4.90 m/s). It should then have a slope of -9.8 m/s², crossing through v = 0 at t = 0.55 sec and ending at v = -0.98 m/s at t = 0.65 sec.

(c) This gives t = 2.5 s. Since the ball rises for 2.5 s, the time to fall is 2.5 s.

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Answer:

0 km/h vertically, horizontally yet 350 km/h

Explanation:

The bundle was inside the plane, so it start a free fall movement from rest, i.e., 0km/h, but of course the speed increases with gravity, at 9,8m/s^2 over time, but since here it is asked the vertical speed at the beginning, it is 0 km/h.

Answer:

4.933m/s

Explanation:

the wagon has a weight of 35.1kg*9.81m/s2 = 343.98N

of that weight 343.98N*sin(18.3)=108N are parallel to the hill and oposit tothe tension of the rope.

then, the force that is moving the wagon is 125N-108N=17N

F=m*a then 17N=35.1kg*a

a=0.4843 m/s2

we have two equations

[tex]v=a.t\\x=v.t+\frac{1}{2} . a . t^{2}[/tex]

then

[tex]x=a.t^{2} +\frac{1}{2}. a t^{2} \\x=\frac{3}{2}. a t^{2}[/tex]

[tex]t=\sqrt{\frac{2x}{3a} }[/tex]

t=10.188s

[tex]v=a.t[/tex]

v=4.933 m/s

Explanation:

Given that,

Mass of the object, m₁ = 23 kg

Acceleration of this object, a₁ = 57 m/s/s

Mass of another object, m₂ = 23 kg

We need to find the acceleration of another object. It can be calculated using second law of motion as :

[tex]F=ma[/tex]

[tex]a=\dfrac{F}{m}[/tex]

Here, F is same. So,

[tex]\dfrac{a_1}{a_2}=\dfrac{m_2}{m_1}[/tex]

[tex]a_2=\dfrac{a_1m_1}{m_2}[/tex]

[tex]a_2=\dfrac{57\times 23}{23}[/tex]

[tex]a_2=57\ m/s/s[/tex]

So, another object will create same acceleration as 57 m/s/s. Hence, this is the required solution.

Final answer:

The same force that accelerates an object of 23 Kg at 57 m/s² would also accelerate another object of mass 23 Kg at 57 m/s², as per Newton's second law of motion.

Explanation:

If a certain force accelerates an object of mass 23 Kg at 57 m/s², the same force would produce the same acceleration on another object of mass 23 Kg. This is because acceleration is directly proportional to force and inversely proportional to mass, according to Newton's second law of motion, which can be expressed as F = ma. Since both the force and mass are the same in this scenario, the acceleration would also be the same, which is 57 m/s².

Answer:

The acceleration from it's legs is [tex]a=138.20\frac{m}{s^{2} }[/tex]

Explanation:

Let's order the information:

Initial height: [tex]y_{i}=0m[/tex]

Final height: [tex]y_{f}=2.26m[/tex]

The bush accelerates from [tex]y_{i}=0m[/tex] to [tex]y_{e}=0.16m[/tex].

We can use the following Kinematic Equation to know the velocity at [tex]y_{e}[/tex]:

[tex]v_{f}^{2} = v_{e}^{2} - 2g(y_{f}-y_{e})=2ay_{e}[/tex]

where g is gravity's acceleration (9.8m/s). Since [tex]v_{f}=0[/tex],

[tex]2g(y_{f}-y_{e}) = v_{e}^{2}[/tex]

⇒ [tex]v_{e}=6.41\frac{m}{s}[/tex]

Working with the same equation but in the first height interval:

[tex]v_{e}^{2} = v_{i}^{2} + 2(a-g)(y_{e}-y_{i})[/tex]

Since [tex]v_{i}=0[/tex] and [tex]y_{i}=0[/tex],

[tex]v_{e}^{2} = 2(a-g)y_{e}[/tex]

⇒[tex]a-g=\frac{v_{e}^{2}}{2y_{e}}[/tex]

⇒[tex]a=\frac{v_{e}^{2}}{2y_{e}}+g[/tex] ⇒ [tex]a=138.20\frac{m}{s^{2} }[/tex]

Answer:

yes

Explanation:

because when you slow down, the resistance slows with the speed.

Answer:

Yes, air resistance decrease with speed

Explanation:

Air resistance is a kind of fluid friction that acts when objects flow through fluids. It is affected by the velocity of moving objects and the area of the objects. When an object moves with a greater velocity the air resistance acting on them will be high, so the speed decreases.

Fluid friction acts in fluids namely liquids and gases. In liquids the friction is called buoyancy and in gases it is called air resistance or drag. When two objects having the same mass but different area move through air, the object with larger area will have less velocity compared with the other object.

Answer:

[tex]v = 33.66 m/s[/tex]

Explanation:

Let hockey puck is moving at constant speed v

so here we have

[tex]d = vt[/tex]

so time taken by the puck to strike the wall is given as

[tex]t = \frac{58.2}{v}[/tex]

now time taken by sound to come back at the position of shooter is given as

[tex]t_2 = \frac{58.2}{340}[/tex]

[tex]t_2 = 0.17s[/tex]

so we know that total time is 1.9 s

[tex]1.9 = t + t_2[/tex]

[tex]1.9 = t + 0.17[/tex]

[tex]1.9 - 0.17 = t[/tex]

[tex]t = 1.73 s[/tex]

now we have

[tex]1.73 = \frac{58.2}{v}[/tex]

[tex]v = 33.66 m/s[/tex]

Answer:

  a.  17.0 m²

Explanation:

The product of the two dimensions is 16.99013 m². The least precise contributor has 3 significant figures, so the most precise result available is one with 3 significant figures: 17.0 m².

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Additional comment

Given that each measurement may be in error by 1/2 of a least-significant digit, their product can be as little as 16.9700025, or as great as 17.0102625. This amounts to 16.9901325 ± 0.0201300. The value 17.0 suggests a range from 16.95 to 17.05, which exceeds the actual range possible with the given measurements. On the other hand, a 4 significant-figure value (16.99) suggests a much smaller range in the product than there may actually be: (16.985, 16.995)

The car will pass the truck after 20 seconds.

To find when the car will pass the truck, we need to determine the time it takes for the car to reach a speed of 25 m/s and the distance traveled by the truck in that time.

The car is accelerating at a rate of 1.25 m/s². Using the equation v = u + at, where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration, and t is the time, we can solve for t: 25 m/s = 0 m/s + 1.25 m/s² * t.

Simplifying the equation, we get t = 20 seconds. Therefore, the car will pass the truck after 20 seconds.

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Answer:

[tex]v_o = 2761 m/s[/tex]

Explanation:

Since there is no external Force or Torque on the system of bullet and rod

So we can say that total angular momentum of the system will remain conserved about its center

So we will have

[tex]mv_o\frac{L}{2}sin\theta = (I_{rod} + I_{bullet})\omega[/tex]

here we know that

[tex]I_{rod} = \frac{mL^2}{12}[/tex]

[tex]I_{rod} = \frac{4.10\times 0.70^2}{12} [/tex]

[tex]I_{rod} = 0.167 kg m^2[/tex]

[tex]I_{bullet} = mr^2[/tex]

[tex]I_{bullet} = (0.003)(0.35^2)[/tex]

[tex]I_{bullet} = 3.675 \times 10^{-4} kg m^2[/tex]

[tex]\omega = 15 rad/s[/tex]

[tex]\theta = 60 [/tex]

now we have

[tex]0.003(v_o)(0.35)sin60 = (0.167 + 3.675 \times 10^{-4})15[/tex]

[tex]v_o = 2761 m/s[/tex]

This physics question is about calculating the speed of a bullet before it strikes a rotating rod, based on the conservation of angular momentum. After setting up the equations for initial and final angular momentum, by equating these two, we can find the speed of the bullet just before it hits the rod.

This is a physics problem regarding the conservation of angular momentum during a collision. We are given a scenario where a bullet impacting and lodging into a rotating rod which is initially at rest. According to the conservation of angular momentum, the initial angular momentum prior to the collision should equal the angular momentum after the collision, if no external torque is acting on the system.

The initial angular momentum (just before the collision) is the product of the bullet's mass, speed, and distance from the axis of rotation (which is half of the rod's length), and the cos(θ), where θ is the angle the bullet's path makes with the rod. In this case, angular momentum is mbrvb cos(θ)r

The final angular momentum (just after the collision) is the moment of inertia of the system (bullet plus rod) times the ensuing angular velocity, which is (mb(r^2) + (1/12)M(r^2))ω

By setting the initial and the final angular momentum equations equal to each other and arranging for vb, we will find the speed of the bullet just before impact.

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Answer:

Technician A says that this is because the front wheel cylinders are closer to the master cylinder than are the rear wheel cylinders.

Explanation:

Since the position of cylinder is near the front wheel so the normal force of the front wheel is more than the normal force on the rear wheel

as we know that the center of mass of the wheel is shifted towards the front wheel as the balancing is done with reference to its center of mass

so we will have

[tex]N_1 d_1 = N_2 d_2[/tex]

also we have

[tex]N_1 + N_2 = W[/tex]

so here we can say

friction force on front wheel will be more

[tex]F_f = \mu N_1[/tex]

so here front wheel has to do more work to stop the vehicle

Technician B is correct; the front brakes do more work because the vehicle's weight shifts to the front during braking due to dynamic load transfer. This is a physics principle related to force and motion, rather than the hydraulic pressure distribution which is equal for all brakes due to Pascal's principle.

Technician B suggests that the front brakes do more work than the rear brakes because the weight of a vehicle shifts to the front during a stop. This is correct, as deceleration causes the vehicle's weight to transfer to the front due to inertia, resulting in greater pressure on the front brakes. This phenomenon is known as weight transfer or dynamic load transfer. The claim by Technician A that proximity to the master cylinder affects braking force is not accurate, as hydraulic systems utilize Pascal's principle to ensure equal pressure distribution throughout the brake fluid regardless of the distance from the master cylinder.

Answer:

No the distance traveled in last 0.1 s is not same as that the distance traveled in first 0.1 s

so it will cover more distance in last 0.1 s then the distance in first 0.1 s

Explanation:

As we know that when stone is dropped from the diving board then its velocity at the time of drop is taken to be ZERO

so here we can say that its displacement from the top position in next 0.1 s is given as

[tex]d_1 = v_y t + \frac{1}{2}at^2[/tex]

[tex]d_1 = 0 + \frac{1}{2}(9.81)(0.1)^2[/tex]

[tex]d_1 = 0.05 m[/tex]

Now during last 0.1 s of its motion the stone will attain certain speed

so we will have

[tex]d_2 = v_y(0.1) + \frac{1}{2}(9.81)(0.1)^2[/tex]

[tex]d_2 = 0.1 v_y + 0.05 m[/tex]

so it will cover more distance in last 0.1 s then the distance in first 0.1 s

Final answer:

The distance a rock travels during a 0.1-second interval while falling will vary because of acceleration due to gravity. Early in its fall, it will travel less distance, and just before impact, it will cover more distance due to increased speed.

Explanation:

The distance traveled by a rock in free fall will not be the same in two intervals of time if these intervals occur at different stages of the fall, because the rock is accelerating due to gravity.

Near the top of its flight, it will have just started to accelerate, so it will cover a smaller distance in the first 0.1-second interval. However, just before the rock hits the water, it will have been accelerating for the entire duration of the fall, meaning it will be traveling much faster and will cover a greater distance in the last 0.1-second interval before impact.

Answer:

C) about 1/10 as great.

Explanation:

We use the relation between Impulse, I, and momentum, p:

[tex]I=\Delta p\\ F*t=m(v_{f}-v{o})\\ \\[/tex]

[tex]F=m(v_{f}-v{o})/t=-mv{o}/t\\ \\[/tex]     the final speed is zero

We can see that the average Force is inversely proportional to the time, so if the time is 10 times bigger, the average Force is 1/10 as great

Average force body experience is about 1/10 as great as  when the legs are kept stiff

from Newtons second law, force. f is the product of mass, m and acceleration, a

f = m * a

a = ( v - u ) / t

f = m * ( v- u ) / t

so force is inverrsely related to time, making the time of the impact 10 times as great as stiff-legged landing impiles mathematically as follows:

for stiff-legged

f1 = m * ( v - u ) / t

for bent knees

f2 = m * ( v - u ) / 10t

f2 = f1 / 10

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Answer:

An annular Solar Eclipse

Explanation:

Solar eclipse is an event that occurs naturally on Earth when the moon in its orbit is positioned between the Earth and the Sun.Solar Eclipse can be total ,partial or annular.In the total solar eclipse, the moon completely covers the sun where as in the annular solar eclipse the moon covers the center of the Sun leaving outer edges of the Sun to be visible forming the ring of fire.In partial solar eclipse the Earth moves through the lunar penumbra as the moon moves between Earth and Sun.The moon blocks only some parts of the solar disk.Annular solar eclipse happens during new moon and the moon is at its farthest position from the Earth called Apogee.

A solar eclipse occurs when the moon blocks the sun. It can either be a total eclipse, where the sun is completely covered, or a partial or annular eclipse when the new moon is too far to entirely cover the sun.

A solar eclipse occurs when the moon moves between the earth and the sun, blocking out sunlight and casting a shadow. This can either result in a total eclipse, where the full face of the sun is covered, or a partial eclipse when part of the sun is still visible. A total eclipse only happens when the moon is close enough to the earth to totally cover the sun. When the new moon is too far from the earth to completely cover the sun, the eclipse can be either partial or become what's known as an 'annular' eclipse. In an annular solar eclipse, the moon is located too far from the Earth to completely cover the sun's disk, resulting in a ringlike appearance.

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Answer:

[tex]2LiBr+Pb(NO_3) = PbBr_2+ 2LiNO_3[/tex]  is a double displacement reaction

Explanation:

[tex]2LiBr+PbNO_3 = PbBr_2+2LiNO_3[/tex]

                                                       It is of the form [tex]AX+BY = AY+BX[/tex]

                                                       [tex]Fe+2HCl = FeCl_2+H_2[/tex]

Both oxidation and reduction occur in this reaction. Fe gets oxidized and H gets reduced.

                                          [tex]Fe = Fe^(2+)+2e^-[/tex]

                                          [tex]2H^++2e^- = H_2[/tex]

                                          [tex]CaO+H_2 O = Ca(OH)_2[/tex]

It is an exothermic combination reaction. Calcium reacts with water to produce calcium hydroxide and heat is released in this process.

It is a decomposition reaction where nickel chloride decomposes to nickel and chlorine.

Answer:

18. Double Displacement Reaction

19. Single Displacement Reaction

20. Synthesis Reaction

21. Decomposition Reaction

Explanation:

Double Displacement formula:

AB + CD --> AD + CB

Single Displacement formula:

A + BC --> AC + B

Synthesis formula:

A + B --> AB

Decomposition formula:

AB --> A + B

The work done in stretching a spring in two stages can be estimated by dividing the stretching process into two stages, calculating the average force in each stage, and then calculating the work done in each stage. The total work done is the sum of the work done in each stage.

To calculate the work done in stretching the spring in two stages, we can assume an average force for each stage. The force exerted by the spring at any displacement from its equilibrium length is given by Hooke's Law, F = kx, where 'F' is the force, 'k' is the spring constant, and 'x' is the displacement from equilibrium.

For stretching the spring from 0.3 m to 0.5 m (stage one), the average force exerted would be F_avg1 = k(x2 + x1) / 2 = 22 * ( 0.5 + 0.3 ) / 2 = 8.8 N. The work done can be then calculated as Work1 = F_avg1 * (x2 - x1) = 8.8 * (0.5-0.3) = 1.76 J.

For stretching the spring from 0.5 m to 0.7 m (stage two), the average force would be F_avg2 = k(x2 + x1) / 2 = 22 * ( 0.7 + 0.5 ) / 2 = 13.2 N. The work done in this stage would be Work2 = F_avg2 * (x2 - x1) = 13.2 * (0.7 - 0.5) = 2.64 J.

Therefore, the total work done in stretching the spring in two stages would be Work_total = Work1 + Work2 = 1.76 J + 2.64 J = 4.4 J.

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To estimate the work done in stretching the spring, divide the process into two stages: 0.3 m to 0.5 m and 0.5 m to 0.7 m. The total work done is 4.4 J.

Stage 1: Stretching from 0.3 m to 0.5 m

The initial force (F₁) at 0.3 m is:

The final force (F_mid) at 0.5 m is:

The average force (F_avg₁) for this stage is:

The distance stretched (d1) is:

Thus, the work done (W₁) is:

Stage 2: Stretching from 0.5 m to 0.7 m

The initial force (F_mid) at 0.5 m is:

The final force (F₂) at 0.7 m is:

The average force (F_avg₂) for this stage is:

The distance stretched (d2) is:

Thus, the work done (W₂) is:

Total Work Done in stretching the spring from 0.3 m to 0.7 m is:

Answer:

(a) The x-component of velocity is 31.55 km/h

(b) The y-component of velocity is 44.92 km/hr

Solution:

As per the solution:

The relative position of ship A relative to ship B is 4.2 km north and 2.7 km east.

Velocity of ship A, [tex]\vec{u_{A}}[/tex] = 22 km/h towards South = [tex]- 22\hat{j}[/tex]

Velocity of ship B, [tex]\vec{u_{B}}[/tex] = 39 km/h Towards North east at an angle of [tex]36^{\circ}[/tex] = [tex]\vec{u_{B}} = 39sin36^{\circ} \hat{j}[/tex]

Now, the velocity of ship A relative to ship B:

[tex]\vec{u_{AB}} = \vec{u_{A}} - \vec{u_{B}}[/tex]

[tex]\vec{u_{A}} = - 22\hat{j}[/tex]

[tex]\vec{u_{B}} = 39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}[/tex]

Now,

[tex]\vec{u_{AB}} = - 22\hat{j} +39cos36^{\circ} \hat{i} - 39sin36^{\circ} \hat{j}[/tex]

[tex]\vec{u_{AB}} = 31.55\hat{i} - 44.92\hat{j}[/tex]