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valina [46]
3 years ago
11

The 1350-kg car has a velocity of 24 km/h up the 8-percent grade when the driver applies more power for 18 s to bring the car up

to a speed of 65 km/h. Calculate the time average F of the total force tangent to the road exerted on the tires during the 18 s. Treat the car as a particle and neglect air resistance.
Physics
1 answer:
Brrunno [24]3 years ago
4 0

Answer:

Explanation:

We shall apply concept of impulse to solve the problem  .

Impulse = force x time

impulse = change in momentum

force x time = change in momentum

initial speed u = 24 km/h = 6.67 m /s

final speed v = 65 km/h = 18.05 m /s

change in momentum = m v - mu

= m ( v-u )

= 1350 ( 18.05 - 6.67 )

= 15363 kg m/s

F x 18 = 15363

F = 853.5 N .

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5.5 billion years

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What charge do electrons have?<br>Negative<br>Positive<br>Neutral​
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3 years ago
A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00
vivado [14]

Answer:

7.5 m

Explanation:

v = initial speed of the ball = 8 m/s

\theta = angle of launch = 40° deg

Consider the motion along the vertical direction :

v_{oy} = initial velocity along vertical direction = v Sin\theta = 8 Sin40 = 5.14 m/s

a_{y} = acceleration along vertical direction = - 9.8 m/s²

t = time of travel

y = vertical displacement = - 1 m

Using the kinematics equation

y = v_{oy}t + (0.5)a_{y}t^{2}

- 1 = (5.14)t + (0.5)(- 9.8)t^{2

t = 1.22 sec

Consider the motion along the horizontal direction :

v_{ox} = initial velocity along horizontal direction = v Sin\theta = 8 Cos40 = 6.13 m/s

a_{x} = acceleration along vertical direction = 0 m/s²

t = time of travel = 1.22 sec

x = horizontal displacement = ?

Using the kinematics equation

x = v_{ox}t + (0.5)a_{x}t^{2}

x = (6.13)(1.22) + (0.5)(0)(1.22)^{2

x = 7.5 m

4 0
3 years ago
The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of
PIT_PIT [208]

Answer:

B = 4.1*10^-3 T = 4.1mT

Explanation:

In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:

\Phi_B=S\cdot B=SBcos\alpha        (1)

ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2

S: surface area of the circular plate = π.r^2

r: radius of the circular plate = 8.50cm = 0.085m

B: magnitude of the magnetic field = ?

α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°

You solve the equation (1) for B, and replace the values of the other parameters:

B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT

The strength of the magntetic field is 4.1mT

4 0
3 years ago
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