A boy in a sleigh glides on a 40 angle snow-covered hill. The coefficient of kinetic friction between the surface and the sled is 0.12. The acceleration of the sled is: a. 3.6 m / s^2
b. 0.76 m / s^2
c. 2.3 m / s^2
d. 5.4 m / s^2

Answer:16.59 m

Explanation:

Given

initial horizontal speed of ball(u)=15 m/s

Height of building =6 m

Consider vertical motion first

[tex]h=u_yt+\frac{1}{2}gt^2[/tex]

here initial vertical velocity is zero

[tex]6=0+\frac{1}{2}\times 9.81\times t^2[/tex]

[tex]t=\sqrt{\frac{12}{9.81}}=1.106 s[/tex]

Thus time taken will also be 1.106 s in horizontal motion

[tex]R_x=u_xt+\frac{1}{2}at^2[/tex]

here a=0

[tex]R_x=15\times 1.106=16.59 m[/tex]

The total time that the ball is in the air can be calculated using the formula for free fall, which is 1.10 seconds. Given that time and the constant horizontal speed, we can then calculate the total horizontal distance traveled by the ball, which is 16.5 meters.

To calculate the horizontal distance the ball travels before striking the ground, we must first determine how long the ball is in the air. This time is only affected by the vertical motion, thus we can treat the ball as if it were dropped from the hill with no initial horizontal velocity. The formula used to calculate the time is based on free fall, where t = sqrt((2*h)/g). In this case, h = 6.0 m (the height of the hill) and g = 9.8 m/s² (the acceleration due to gravity). This gives us t = sqrt((2*6)/9.8) = 1.10 s.

Once we have the time of flight, we can calculate the horizontal distance traveled using the horizontal speed of the ball. As the horizontal motion occurs at a constant velocity we can use the formula d = v*t, where v = 15 m/s (the speed the ball is thrown) and t = 1.10 s. This gives us the horizontal distance d = 15*1.10 = 16.5 m.

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Answer:

The no. of electrons is [tex]7.59\times 10^{21}[/tex]

Solution:

According to the question:

The rate at which the charge is delivered is given by:

[tex]\frac{dQ}{dt} = - 0.75[/tex]

Now,

[tex]\int_{0}^{Q}dQ = - 0.75\int_{0}^{27 min} dt[/tex]

[tex]Q = -0.75t|_{0}^{27 min}[/tex]

[tex]Q= -0.75\times 27\times 60 = - 1215 C[/tex]

No. of electrons, n can be calculated from the following relation:

Q = ne

where

e = electronic charge =[tex]1.6\times 10^{- 19} C[/tex]

Thus

[tex]n = \frac{Q}{e}[/tex]

[tex]n= \frac{1215}{1.6\times 10^{- 19}}[/tex]

[tex]n = 7.59\times 10^{21}[/tex]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

Solution:

As per the question:

Velocity of moving frame w.r.t original frame [tex]v_{m}[/tex] 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = [tex]\frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}[/tex]

X = [tex]\frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}[/tex]

X = [tex]8.07 c[/tex]

Now,

Y = y = - 2

Z = z = 3

Now,

[tex]t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}[/tex]

[tex]t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s[/tex]

To find the position, velocity, and acceleration of the snowboarder at t = 12 seconds, evaluate the given displacement equation and its first and second derivatives at t = 12 seconds, then round the final answers to one decimal place.

To determine the position, velocity, and acceleration of the snowboarder at t = 12 seconds, we need to evaluate the given displacement function x(t) = 0.5t³ + t² + 2t and its derivatives at that specific time value.

Position at t = 12 seconds

By substituting t = 12 into the displacement function, we get:
x(12) = 0.5(12)³ + (12)² + 2(12), which can be calculated to give the position x.

Velocity at t = 12 seconds

The velocity v(t) is the first derivative of the displacement function, v(t) = 1.5t² + 2t + 2. Evaluate v(12) to find the velocity at t = 12 seconds.

Acceleration at t = 12 seconds

Acceleration a(t) is the second derivative of the displacement function, a(t) = 3t + 2. Evaluate a(12) to find the acceleration at t = 12 seconds.

Using these steps, you can calculate the exact values and round them to one decimal place as asked in the question.

Answer:

a) The average velocity is v = (60 km/h ; 0)

b) The average speed is 60 km/h

Explanation:

The velocity is a vector that has a magnitude and direction. The average speed is the distance traveled over time without taking into account the direction of the motion.

a)The average velocity is calculated as the displacement over time:

v = Δx/Δt

where

v = velocity

Δx = final position - initial position = traveled distance relative to the center of the reference system.

Δt = final time - initial time (initial time is usually = 0)

We know that the displacement is 84 km but we do not know the time. It can be calculated from the two parts of the trip.

In part 1:

v = 45 km/h = 42 km / t

t = 0.93 h

In part 2:

v = 90 km/h = 42 km / t

t = 42 km / 90 km/h

t = 0.47 h

The time of travel is 0.47 h + 0.93 h = 1.4 h

The average velocity will be:

v = 84 km / 1.4 h = 60 km/h

Expressed as a vector in a 2-dimension plane:

v = (60 km/h; 0)

b) The average speed is calculated as the distance traveled over time. Note that in this case, the distance is equal to the displacement since the direction of the motion is always in one direction. But if the direction of the second part of the trip would have been the opposite to the direction of the first part, the displacement would have been 0 (final position - initial position = 0, because final position = initial position), then, the average velocity would have been 0. In change, the average speed would have been the distance traveled (84 km, 42 km in one direction and 42 km in the other) over time.

Then:

average speed = 84 km / 1.4 h = 60 km/h

c) see attached figure.    

Answer:

(a) -3.173 m/s^2

(b) 3.94 s

(c) 2.47 s

Explanation:

initial velocity, u = 25 m/s

final velocity, v = 0

distance, s = 98.5 m

(a) Let a be the acceleration of the car

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

0 = 625 + 2 x a x 98.5

a = -3.173 m/s^2

(b) v = 12.5 m/s

u = 25 m/s

a = - 3.173 m/s^2

Let the time is t.

Use first equation of motion

v = u + a t

12.5 = 25 - 3.173 t

t = 3.94 s

(c) s = 98.5 / 2 = 49.25 m

u = 25 m/s

a = - 3.173 m/s^2

Let the time be t.

Let v be the velocity at this distance.

Use third equation of motion

[tex]v^{2}=u^{2}+2as[/tex]

[tex]v^{2}=25^{2}+2\times {-3.173}\ times 49.25tex]

v = 17.17 m/s

Use first equation of motion

v = u + at

17.17 = 25 - 3.173 x t

t = 2.47 s

Answer:

[tex]f_p = 5000 N[/tex]

Explanation:

GIVEN DATA:

pressure ratio = length ratio

force = 5 N

scale = 1:10

velocity = 1.5 m/s

[tex]B_p = 20 m[/tex]

[tex]h_p = 2m[/tex]

As pressure ratio = length ratio so we have

[tex]\frac{p_m}{p_p} =\frac{l_m}{l_p} =\frac{1}{10}[/tex]

[tex]\frac{f_m *A_m}{f_p *A_p} ==\frac{1}{10}[/tex]

[tex]\frac{f_m}{f_p} * \frac{A_p}{A_m} = \frac{1}{10}[/tex]

[tex] \frac{f_m}{f_p} * \frac{b_p*h_p}{b_m*h_m} =\frac{1}{10}[/tex]

[tex]5 * \frac{1}{\frac{1}{10}} *\frac{1}{\frac{1}{10}} = \frac{f_p}{10}[/tex]

solving F_p

[tex]f_p = 5000 N[/tex]

Answer:

B) 33 m C) 27 m

Explanation:

considering that the two pucks are sliding toward each other we can understand that they are on a collision course.

Since the total distance between them is 26 m, the common sense dictates that the distance traveled by each puck must be less than 26 m regardless of the speed of the two pucks.

so the options B) 33 m and C) 27 m are definitely wrong since they are greater than 26 m.

We can also easily find the distance traveled by each pucks also.

let [tex]v_{A}[/tex] and [tex]v_{B}[/tex] be the velocity of the pucks A and B respectively

[tex]v_{A}t = 26- v_{B}t\\\\2.30t = 26 - 3.90t\\\\6.2t = 26\\\\t = \frac{26}{6.2} \\\\x_{A} =v_{A}t\\\\x_{A} =2.3 \times \frac{26}{6.2}\\\\x_{A} =9.65\\\\x_{B} =v_{B}t\\\\x_{B} =3.9 \times \frac{26}{6.2}\\\\x_{B} =16.35\\[/tex]

Answer:

Truck's speed = 5.21 m/s

Car's speed = 20.2 m/s

Explanation:

Given:

Mass of truck = M = 1650 kg

Speed of the truck initially = U = 15 m/s

Mass of the car = m = 779 kg

Initial speed of the car =u = 0

From the momentum conservation, Total initial momentum = Total final momentum.

M V+m U = M V +m v

⇒ (1650)(15) + 779×0 = (1650)V + 779 v

⇒ 24750 = 1650 V+779 v →(1)

Since the collision is elastic, relative velocity of approach = relative velocity of separation. 15 = v - V

⇒ v =V + 15; This is now substituted in the equation(1) above.

24750 = 1650 V + (799) (V+15)

⇒ 24750 = 1650 V + 799 V + 11985

⇒ 2449 V = 12765

⇒ Final velocity of the truck = [tex]\frac{12765}{2449}[/tex] = 5.21 m/s

Final velocity of the car = v = V+15 = 5.21 + 15 = 20.2 m/s

Answer:

[tex]X - Xo = 54m[/tex]

k = 1/18

Explanation:

Data:

a = -k[tex]t^{2}[/tex][tex]\frac{m}{s^{2} }[/tex]

to = 0s    Vo = 12m/s

t = 6s the particle chage it's moviment, so v = 0 m/s

We know that acceleration is the derivative of velocity related to time:

[tex]a = \frac{dV}{dT}[/tex]

rearranging...

[tex]a*dT = dV[/tex]

Then, we must integrate both sides:

[tex]\int\limits^f_i {dV} \, dV =-k \int\limits^f_i {t^{2} } \, dT[/tex]

[tex]V - Vo = -k\frac{t^{3} }{3}[/tex]

V = 0 because the exercise says that the car change it's direction:

[tex]0 - 12 = -k\frac{6^{3} }{3}[/tex]

k = 1/6

In order to find X - Xo we must integer v*dT = dX

[tex]V - Vo = -k\frac{t^{3} }{3}[/tex]

so...

[tex](Vo -k\frac{t^{3} }{3})dT = dX[/tex]

[tex]\int\limits^f_i {dX} \, dX = \int\limits^f_i {Vo -k\frac{t^{3} }{3} } \, dT[/tex]

integrating...

[tex]X - Xo = Vot -k\frac{t^{4} }{12}[/tex]

[tex]X - Xo = 12*6 -\frac{1}{6}* \frac{6^{4} }{12}[/tex]

X - Xo = 54m

Answer:

a) The car doesn't hit the barrier because after he sees the lights of a barrier he only travels for 27.95m, enough to miss the barrier. b) The maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s or 75.6km/hr.

Explanation:

a)

In order to solve this problem we must first do a drawing of the situation for us to visualize it better. (See picture attached).

As you may see on the drawing, the car still travels some distance when the driver notices the lights of the barrier. This distance is calculated for a constant velocity V:

x=Vt

the car has a velcity of 60km/h which is equivalent to:

[tex]\frac {60km}{hr} * \frac{1000m}{1km}*\frac{1hr}{3600s}=16.667m/s[/tex]

so on the first 0.80s the car travels a distance of:

x=(16.667m/s)(0.8s)=13.333m

Next, the car breaks, so we can say it moves with a constant acceleration of -9.5m/s, so the distance is found by using the following formula:

[tex]x=\frac {V_{f}^{2}-V_{0}^{2}}{2a}[/tex]

We know the final velocity will happen when the car stops, so the final velocity is zero, leaving us with:

[tex]x=\frac {-V_{0}^{2}}{2a}[/tex]

so we can substitute the provided values to find the distance traveled by the car during this time.

[tex]x=\frac{-(16.667m/s)^{2}}{2(-9.5m/s^{2})}[/tex]

which yields:

x=14.62m

so in total the car traveled:

[tex]x_{tot}=13.333m+14.62m = 27.95m[/tex]

Which is not enough for the car to hit the barrier.

b)

In order to solve this part of the problem, we must combine the two equations we got on the previous part to find a single equation that will represent the total displacement of the car:

[tex]x=vt-\frac{v^{2}}{2a}[/tex]

so we can now substitute the known values so we get:

[tex]40=0.8v-\frac{v^{2}}{2(-9.5)}[/tex]

which simplifies to:

[tex]40=0.8v+0.05263v^{2}[/tex]

this is a quadratic equation so it can be solved by using the quadratic formula, but first we must rewrite it in standard form, so we get:

[tex]0.05263v^{2}+0.8v-40=0[/tex]

now we can use the quadratic formula:

[tex]v=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex]

So we can substitute the given values:

[tex]v=\frac{-0.8\pm\sqrt{(0.8)^{2}-4(0.05263)(-40)}}{2(0.05263)}[/tex]

which returns two answers:

v=21m/s  and   v=-36.20m/s

We take the positive answer since is the one that represents a moving towards the right side of the drawing.

So the maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s

The electrostatic force between an electron and a proton separated by 0.1mm, calculated using Coulomb's law, is approximately 2.3 × 10^-20N, and it is attractive due to the opposite charges.

The electrostatic force between charged particles can be calculated using Coulomb's law: F = k(Q1*Q2)/r², where F is the force, k is Coulomb's constant (~8.99 * 10^9 Nm^2/C^2), Q1 and Q2 are the charges, and r is the separation in meters. An electron and a proton have charges of -1.6*10^-19 C and +1.6*10^-19 C respectively. Plugging these values into Coulomb's law with a separation of 0.1 mm or 0.1*10^-3 meters, we find that the force is approximately 2.3 × 10^-20N, and it is attractive because the charges are opposite.

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"The correct answer is c. [tex]\(2.3 \times 10^{-18}\)[/tex] N, attractive.

To find the electrostatic force between an electron and a proton separated by a distance, we use Coulomb's law, which is given by:

[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]

The charge of an electron is approximately [tex]\(-1.602 \times 10^{-19}\) C[/tex]and the charge of a proton is approximately [tex]\(+1.602 \times 10^{-19}\) C[/tex]. The separation distance is given as 0.1 mm, which we need to convert to meters: [tex]\(0.1 \times 10^{-3}\)[/tex] m or [tex]\(1 \times 10^{-4}\)[/tex] m.

Plugging in the values, we get:

[tex]\[ F = (8.9875 \times 10^9) \frac{(1.602 \times 10^{-19})(1.602 \times 10^{-19})}{(1 \times 10^{-4})^2} \] \[ F = (8.9875 \times 10^9) \frac{2.5664 \times 10^{-38}}{1 \times 10^{-8}} \] \[ F = (8.9875 \times 10^9) (2.5664 \times 10^{-30}) \] \[ F = 2.3017 \times 10^{-19} \][/tex]

Since the force is attractive (opposite charges attract), the magnitude of the force is [tex]\(2.3 \times 10^{-19}\) N[/tex], but we need to express it in the format given in the options, which is [tex]\(2.3 \times 10^{-18}\) N[/tex].

Answer:

3.22 s

Explanation:

The ball would describe a projectile motion, where the horizontal velocity will remain constant, as there are no forces that act on the x-axis, and the vertical velocity will vary because of gravity in the following way:

1. First, the ball will go up, but the vertical velocity will decrease until it has a value of 0.

2. After the vertical velocity has reached the value of 0, the ball will start to fall, with the vertical velocity increasing because of gravity.

You need to know that the time that the ball's verical velocity takes to reach 0 is exactly the same that it takes to go from 0 to its original vertical velocity:

[tex]a = \frac{v_f - v_o}{t}\\t = \frac{v_f - v_o}{a}[/tex]

And not only the time will be the same, but also the distance traveled. Therefore, we can conclude that the time that the ball remain in the air is simply two times the time it takes for the ball to desacelerate:

[tex]t_{desaceleration} = \frac{v_f - v_o}{a} = \frac{0m/s - 15.8m/s}{-9.81 m/s^2} =1.61 s\\t_{air} = 2* t_{desaceleration} = 2*1.61s = 3.22 s[/tex]

Answer:19.3 km,[tex]\theta =44.40^{\circ}[/tex] south of east

Explanation:

Given

Hiker treks [tex]30 ^{\circ}[/tex] south of east at a speed of 15 m/s for 30 min and then turns due to west and hikes at speed of 8 m/s for another 20 min

Let position vector of Hiker at the end of 30 min

[tex]r=27000cos30\hat{i}-27000sin30\hat{j}[/tex]

after he turns west so new position vector of hiker is

[tex]r'=27000cos30\hat{i}-27000sin30\hat{j}-9600\hat{i}[/tex]

[tex]r'=13782.68\hat{i}-13500\hat{j}[/tex]

Therefore Displacement is given by |r'|

[tex]|r'|=\sqrt{13782.68^2+13500^2}[/tex]

[tex]|r'|=19,292.8035 m\ or 19.3 km[/tex]

for direction

[tex]tan\theta =\frac{13500}{13782.68}[/tex]

[tex]\theta =44.40^{\circ}[/tex] south of east

Answer:

rock A: [tex]y=90-1/2*g*t^2[/tex]

rock B: [tex]y=30*t-1/2*g*t^2[/tex]

g=9.81m/s^2

Explanation:

Kinematics equation:

[tex]y=y_{o}+v_{oy}*t+1/2*a*t^2[/tex]

in our case the acceleration is the gravity and it has a negative direction.

a=-g

rock A, yo=90m, Voy=0m/s:

[tex]y=90-1/2*g*t^2[/tex]

rock B, yo=0m, Voy=30m/s:

[tex]y=30*t-1/2*g*t^2[/tex]

Explanation:

when a person jumps on the trampoline he stores his potential energy in the form of elastic energy of trampoline. Potential Energy converts into Elastic energy when a person is at the bottom point during stretching of the trampoline. When he again going upwards, Elastic energy is converted to the potential energy of the person. This is the reason why a person goes higher each time. This process goes on until trampoline reaches its elastic limit and finally breaks or get permanently deformed.

So there is a limit up to which a person can be reached.

Final answer:

A person goes higher on a trampoline by landing and pushing off with their feet, making better use of leg strength to transform kinetic to potential energy. There is a maximum height due to energy losses like air resistance and imperfect energy transfers. The sweet spot in tennis rackets illustrates efficient energy transfer with reduced arm jarring.

Explanation:

A person on a trampoline goes higher with each bounce because they can harness the elastic potential energy stored in the trampoline material. When landing on the back or feet, the height reached can differ.

Typically, a person may reach greater heights when landing and launching off their feet because they can make use of the stronger leg muscles to add upward force, thus converting more kinetic energy to gravitational potential energy upon ascent. There is, indeed, a maximum height obtainable, as energy losses due to factors like air resistance and less-than-perfect energy transfer prevent infinite increases in height.

Addressing the professional application, the 'sweet spot' on a tennis racket is an example of an optimal point where energy transfer from the racket to the ball is the most efficient, and vibrating force transmitted to the player's arm is minimal, hence no jarring of the arm.

The physics of motion and energy conservation provides us the information that increasing the initial speed (kinetic energy) of the object would result in a higher ascent, yet this is bounded by practical limitations such as the strength of the jumper and the efficiency of the trampoline material.

For example, to double the impact speed of a falling object, one would have to quadruple the height from which it falls, due to the relationship between gravitational potential energy and kinetic energy.

Answer:

voltage across capacitor [tex]V_C=i\times X_C=0.022\times 0.04=9\times 10^{-4}V[/tex]

Voltage across resistor [tex]V_R=i\times R=0.022\times 2000=80V[/tex]

Explanation:

We have given resistance R = 2000 OHM

Capacitance [tex]C=1000\mu F=0.001F[/tex]

Voltage V = 45 volt

Frequency = 4000 Hz

Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{2\times \pi \times 4000\times 0.001}=0.04ohm[/tex]

Impedance [tex]Z=\sqrt{R^2+X_C^2}=\sqrt{2000^2+0.04^2}=2000ohm[/tex]

Current [tex]i=\frac{V}{Z}=\frac{45}{2000}=0.022A[/tex]

Now voltage across capacitor [tex]V_C=i\times X_C=0.022\times 0.04=9\times 10^{-4}V[/tex]

Voltage across resistor [tex]V_R=i\times R=0.022\times 2000=80V[/tex]

Answer:

E=[tex]k*\frac{q}{21}*u[/tex]

[tex]u=\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]

[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]

Explanation:

q: particle's charge

k: coulomb constant

E=E*u

r=r*u

r=distancia vectorial entre P y S

r=distancia escalar entre P y S

E: Electric field vector

E: magnitud of magnetic field vector

u: unit vector radial

then:

[tex]E=k*q/r^{2}[/tex]

r=r*u

r=P-S=(-1,-2,-4)m

[tex]r^{2}=(Magnitude(P-S))^2=(-1)^2+(-2)^2+(-4)^2=21[/tex]

[tex]r=\sqrt{21}[/tex]

E=[tex]k*\frac{q}{21}*u[/tex]

u=r/r=[tex]\frac{1}{\sqrt{21}} *(-1,-2,-4)m[/tex]

[tex]u_{y}=\frac{-2}{\sqrt{21}} m[/tex]

Answer:

(a) A = [tex]3.90 \AA[/tex]

(b) [tex]A = 4.50 \AA[/tex]

(c) [tex]A = 5.51 \AA[/tex]

(d) [tex]A = 9.02 \AA[/tex]

Solution:

As per the question:

Radius of atom, r = 1.95 [tex]\AA = 1.95\times 10^{- 10} m[/tex]

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = [tex]2\times 1.95 = 3.90 \AA[/tex]

(b) For body centered cubic lattice:

[tex]A = \frac{4}{\sqrt{3}}r[/tex]

[tex]A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA[/tex]

(c) For face centered cubic lattice:

[tex]A = 2{\sqrt{2}}r[/tex]

[tex]A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA[/tex]

(d) For diamond lattice:

[tex]A = 2\times \frac{4}{\sqrt{3}}r[/tex]

[tex]A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA[/tex]

Answer:

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Explanation:

u = Object distance

v = Image distance

f = Focal length = 35

m = Magnification = 2.5

[tex]m=-\frac{v}{u}\\\Rightarrow 2.5=-\frac{v}{u}\\\Rightarrow v=-2.5 u[/tex]

Lens equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{35}=\frac{1}{u}+\frac{1}{-2.5u}\\\Rightarrow \frac{1}{35}=\frac{3}{5u}\\\Rightarrow u=21\ cm[/tex]

[tex]v=-2.5\times 21=-52.5\ cm[/tex]

Image distance is -52.5 cm

Image is virtual and forms on the same side of the lens and upright image is formed.

Answer:

ball hit the ground from her feet is 1.83 m far away

Explanation:

given data

speed = 5.3 m/s

angle = 12°

height = 1 m

to find out

how far from her feet ball hit ground

solution

we consider here x is horizontal component and y is vertical component

so in vertical

velocity will be = v sin12

vertical speed u = 5.3 sin 12 = 1.1 m/s downward

and

in horizontal , velocity we know v = 5.3 m/s

so from motion of equation

s = ut + 0.5×a×t²

s is distance t is time a is 9.8

put all value

1 = 1.1 ( t) + 0.5×9.8×t²

solve it we get t

t = 0.353 s

and

horizontal distance is = vcos12 × t

so horizontal distance = 5.3×cos12 × ( 0.353)

horizontal distance = 1.83 m

so ball hit the ground from her feet is 1.83 m far away

Answer:

It will take 33 seconds to stop the car.

Explanation:

Using the first equation of kinematics we have

[tex]v=u+at[/tex]

where

'v' is final speed of object

'u' is initial speed of object

'a' is acceleration of object

't' is time of acceleration of object

Now since it is given that [tex]a=-2km/s^{2}[/tex] since acceleration is negative  and [tex]u=66km/s[/tex]

We know that the object will stop when it's velocity reduces to zero hence in the equation above setting v = 0 we get

[tex]0=66-2\times t\\\\\therefore t=\frac{66}{2}=33seconds[/tex]

Final answer:

Using the kinematic equation for velocity and acceleration, it is calculated that the car, decelerating at a rate of 2 km/s² from an initial speed of 66 km/s, will take 33 seconds to come to a complete stop.

Explanation:

To find out how long it will take the car to come to a complete stop, we need to use the kinematic equation that relates initial velocity, acceleration, and time without displacement:

Final velocity (v) = Initial velocity (u) + (Acceleration (a) × Time (t))

Here, the final velocity (v) is 0 km/s, since the car is coming to a stop, the initial velocity (u) is 66 km/s, and the acceleration (a) is -2 km/s² (negative because it is deceleration).

The equation becomes:

0 = 66 + (-2 × t)
We can solve for t as follows:

0 = 66 - 2t

2t = 66

t = 33 seconds

So it will take the car 33 seconds to come to a complete stop.

The average speed for the entire trip is 83.6 km/h. The average velocity for the entire trip is zero.

The average speed for the entire trip can be calculated by finding the total distance traveled divided by the total time taken. In this case, the student traveled 39.0 km to school and then returned the same distance back home, resulting in a total distance of 78.0 km. The total time taken for the entire trip is the sum of the time taken to drive to school, the time taken to return home, and the time spent finding the report, which is 23.0 min + 23.0 min + 10.0 min = 56.0 min. To convert the total time to hours, divide by 60: 56.0 min / 60 = 0.933 hours. Therefore, the average speed for the entire trip is 78.0 km / 0.933 hours = 83.6 km/h.

The average velocity for the entire trip can be determined by considering both the magnitude and direction of the displacement. In this case, since the student drove in the same direction to school and back home, the displacement is zero, meaning there was no change in position. Thus, the average velocity for the entire trip is also zero.

Answer:

The mass of the steam is 91.2 g

Mass of the steam=91 grams

Explanation:

Given:

When steam is mixed with the ice then the heat loss by the steam will be gained by the ice so there will no overall heat gain or loss during the mixing

So According to question

Let M be the mass of the steam mixed with ice then we have

[tex]M\times2256\times10^3+M\times4186\times(100-33)=0.499\times222\times10^3+0.499\times4186\times(33-0)\\M\times2.58\times10^6=2.35\times10^5\\M=91.2\ \rm g[/tex]

Answer:

the building is 34.408 m tall

Explanation:

given,

initial velocity of brick = 15.1 m/s

at an angle of = 25.7°

vertical component of the velocity

 Vy = 15.1 sin 25.7°            

       = 6.54 m/s                      

we know                                

[tex]s = u t + \dfrac{1}{2} a t^2[/tex]

[tex]s = 6.54\times 3.4 + \dfrac{1}{2}\times -9.8 \times 3.4^2[/tex]

s = -34.408 m

hence, the building is 34.408 m tall .

Answer:

81.42 km/h

Explanation:

t = Time taken for the car to slow down

u = Initial velocity = 100 km/h

v = Final velocity

s = Displacement = 20 m = 0.02 km

a = Acceleration = -6.5 m/s² = -0.0065×60×60×60×60 = -84240 km/h²

Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -84240\times 0.02+100^2}\\\Rightarrow v=81.42\ km/h[/tex]

Speed of the truck at the end of this distance is 81.42 km/h

Using the equation of motion v^2 = u^2 + 2as, and converting the initial speed from km/hr to m/s, the final velocity is approximately 15.22 m/s when deceleration and distance are considered. Converted back to km/hr, this is approximately 54.8 km/hr.

The final speed of the truck can be calculated using one of the equations of motion, specifically v^2 = u^2 + 2as, where 'v' is final velocity, 'u' is initial velocity, 'a' is acceleration (which in this case is -6.50 m/s^2 due to deceleration), and 's' is distance. Convert the initial speed from km/hr to m/s: 100 km/hr = 27.78 m/s.

Then, we can plug the values into the formula: v^2 = (27.78 m/s)^2 - 2(6.50 m/s^2)(20.0 m). Solving the equation gives a final velocity of approximately 15.22 m/s. This speed in km/h is then obtained by converting m/s back, 15.22 m/s * (3600 / 1000) = 54.8 km/h. So, the speed of the truck at the end of the distance is approximately 54.8 km/h.

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Answer:

0.21 m/s/s.

Explanation:

Whenever there is an action force acting on a body, there will be a reaction force.

Here the force with which the astronaut throws the tool is given as 16 N.

Force is measured in newtons and is equal to the rate of change of momentum.

Since the astronaut has a mass, she experience a reaction force. It is given by F = ma, according to Newton's 2nd law.

16 = 75 a

⇒ Acceleration = a = F/m = 16/75 = 0.21 m/s/s

Answer:

Thus the distance traveled horizontally is 116.28 m

Solution:

As per the question:

Distance covered in 26 steps = 20 m

Thus

Distance covered in one step, step size = [tex]\frac{20}{26}[/tex]

Total steps taken = 152

Now, the distance covered in 152 steps = [tex]one step size\times 152[/tex]

The distance covered in 152 steps, D = [tex]\frac{20}{26}\times 152 = 116.92 m[/tex]

The above distances are on a slope of [tex]6^{\circ}[/tex] above horizontal.

Thus the horizontal component of this distance is given by:

[tex]d_{H} = Dcos6^{\circ}[/tex]

where

[tex]d_{H}[/tex] = horizontal component of distance, D

[tex]d_{H} = 116.92cos6^{\circ} = 116.28 m[/tex]

Final answer:

The volume of the air bubble just before it reaches the surface is 3.32 cm^3.

Explanation:

To find the volume of the air bubble just before it reaches the surface, we can use the ideal gas law equation: PV = nRT. Since the pressure and amount of gas remain constant, we can rewrite the equation as V/T = k, where V is the volume, T is the temperature, and k is a constant.

Using the given temperatures at the bottom and top of the lake (5.9°C and 16.0°C) and the initial volume of the bubble (1.22 cm^3), we can set up the following equation:

(1.22 cm^3) / (5.9°C) = V / (16.0°C).

Solving for V, the volume of the bubble just before it reaches the surface, we get:

V = (1.22 cm^3)(16.0°C) / (5.9°C) = 3.32 cm^3.

To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. If both charges are positive, the net force on a third charge placed on the x-axis will never be zero.

a) To find the location on the x-axis where a third charge can be placed so that the net force on it is zero, we need to consider the forces exerted by the two charges. The net force will be zero when the two forces are equal in magnitude but opposite in direction. Using Coulomb's law, we can calculate the force between the first charge and the third charge at an unknown position x3. We can then set that force equal to the force between the second charge and the third charge at the same position x3, and solve for x3.

b) If both charges are positive, the net force on a third charge placed on the x-axis will never be zero. Positive charges repel each other, so the forces will always be in the same direction.