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8090 [49]
2 years ago
15

A car drives 26.8 m/s south. It

Physics
2 answers:
valina [46]2 years ago
7 0

The time for the car to drive directly south is determined as 7.15 s.

<h3>Time for the car to drive directly west</h3>

The time for the car to drive directly south is determined by applying the concept of slope.

slope = Δy/Δx

a = Δv/Δt

Δt = Δv/a

Δt = (26.8)/(3.75)

Δt = 7.15 s

Thus, the time for the car to drive directly south is determined as 7.15 s.

Learn more about acceleration here: brainly.com/question/605631

#SPJ1

anyanavicka [17]2 years ago
4 0

Answer:

16.96

Explanation:

ax=-3.24

ay=1.58

uiy=-26.8

uix=0.0.0

ufy=0

0=-26.8+1.58t

t=16.96 sec

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What is the name for the change in a stars spectrum when it moves wawy from the earth
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When a source of light moves away from you, you see the characteristic lines in its spectrum move toward slightly longer wavelengths.  Lines in the visible part of the spectrum move toward the red end.

When a source of light moves toward you, you see the characteristic lines in its spectrum move to slightly shorter wavelengths.  Lines in the visible part of the spectrum move toward the violet end.

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8 0
4 years ago
Read 2 more answers
A car is traveling at 108 km/h, stuck behind a slower car. Finally the road is clear and the car pulls over to make a pass. The
mezya [45]

Answer:

The average acceleration of the car is 2.143 meters per square second.

Explanation:

Let assume that car accelerates uniformly, in that case, we can obtain the value of acceleration by using the following equation of motion:

v = v_{o}+a\cdot t

Where:

v_{o} - Initial velocity, measured in meters per second.

v - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t - Time, measured in seconds.

Now, we clear acceleration within expression:

a = \frac{v-v_{o}}{t}

Initial and final velocities are now converted from kilometers per hour into meters per second:

v_{o} = \left(108\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

v_{o} = 30\,\frac{m}{s}

v = \left(135\,\frac{km}{h} \right)\cdot \left(1000\,\frac{m}{km} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

v = 37.5\,\frac{m}{s}

If we know that t = 3.5\,s, then, the average acceleration of the car is:

a = \frac{37.5\,\frac{m}{s}-30\,\frac{m}{s} }{3.5\,s}

a = 2.143\,\frac{m}{s^{2}}

The average acceleration of the car is 2.143 meters per square second.

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3 years ago
Law of motion that says for every action there is an equal and opposite
hodyreva [135]

Answer: Newton's third law

Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.

Explanation:

4 0
3 years ago
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A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc
____ [38]

Answer:

Part a)

a = 0.36 m/s^2

Part b)

a = -1.29 m/s^2

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

F_x = Fcos25

F_x = 80 cos25 = 72.5 N

F_y = Fsin25

F_y = 80 sin25 = 33.8 N

now in vertical direction we can use force balance for the box to find the normal force on it

F_n + F_y = mg

F_n = (25)(9.81) - 33.8

F_n = 211.45 N

now kinetic friction on the box opposite to applied force due to rough floor is given as

F_k = \mu F_n

F_k = (0.300)(211.45) = 63.44 N

now the net force on the box in forward direction is given as

F_{net} = F_x - F_k

F_{net} = 72.5 - 63.44 = 9.065 N

now the acceleration of the box is given as

a = \frac{F_{net}}{m}

a = \frac{9.065}{25} = 0.36 m/s^2

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

F_x = 72.5 N

F_y = 33.8 N

now force balance perpendicular to inclined plane is given as

F_n + F_y = mgcos\theta

F_n = (25)(9.81)cos10 - 33.8 = 207.7 N

now the friction force opposite to the motion on the box is given as

f_k = \mu F_n

f_k = (0.300)(207.7) = 62.3 N

now the net pulling force along the inclined plane is given as

F_{net} = F_x - F_k - mgsin10

F_{net} = 72.5 - 62.3 - (25)(9.81)sin10

F_{net} = -32.38 N

now the box will decelerate and it is given as

a = \frac{F_{net}}{m}

a = \frac{-32.38}{25} = -1.29 m/s^2

7 0
4 years ago
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