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2 years ago

PLEASE ANSWER 50 points

Physics

1 answer:

Evgen [1.6K]2 years ago

6 0

Increase the force the engine provides and decrease the mass of the car

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car is moving at 40 m/s. At 10 meters the driver spots a deer on the road and instantly steps on the brakes. If the car is 400 k

Mice21 [21]

Answer:

32000 N

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Mass (m) of car = 400 Kg

Force (F) =?

Next, we shall determine the acceleration of the the car. This can be obtained as follow:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Acceleration (a) =?

v² = u² + 2as

0² = 40² + (2 × a × 10)

0 = 1600 + 20a

Collect like terms

0 – 1600 = 20a

–1600 = 20a

Divide both side by –1600

a = –1600 / 20

a = –80 m/s²

The negative sign indicate that the car is decelerating i.e coming to rest.

Finally, we shall determine the force needed to stop the car. This can be obtained as follow:

Mass (m) of car = 400 Kg

Acceleration (a) = –80 m/s²

Force (F) =?

F = ma

F = 400 × –80

F = – 32000 N

NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.

7 0

2 years ago

Hannah walks 0.30 km to class in 5.0 min. what is her average speed in m/s?

OverLord2011 [107]

Answer:

1.0 m/s

Explanation:

First, convert to SI units.

0.30 km × (1000 m / km) = 300 m

5.0 min × (60 s / min) = 300 s

Speed is distance divided by time:

300 m / 300 s = 1.0 m/s

3 0

2 years ago

3 + 2 ∙ 4 = 3 + (2 ∙ 4)

Rzqust [24]

Answer:

11 = 11

Explanation:

6 0

2 years ago

Read 2 more answers

The near point of an eye is 48.5 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan

Masteriza [31]

Answer:

The focal length of the lens should be -51.5 cm (a concave lens).

Explanation:

The purpose of the lens is to make objects at 48.5 cm appear at the healthy near point. The healthy near point is 25.0 cm.

We use the lens formula

$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$

where f = focal length, u = object distance and v = image distance.

In this case, u = 48.5 cm and v = -25.0 cm.

v is negative because the image is virtual an not real. (Here, we are using the real-is-positive sign convention)

$\dfrac{1}{f} = \dfrac{1}{48.5} + \dfrac{1}{-25.0} = -\dfrac{23.5}{1212.5}$

$f = -51.5$

The negative sign indicates the lens is concave.

3 0

2 years ago

A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t

Pie

Answer:

$I = \frac{mvb}{6}$

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

$L = I$ ω .... (1)

moment of inertia of rod rotating about its center of length b

$I = \frac{ mb^2}{12}$ ........ .(2)

using v = ωr

where w is angular velocity

and r is radius of rod which is equal to b

so we get 2v = ωb

ω = 2v/b ................. (3)

here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

$L = \frac{mb^2}{12}$ × $\frac{2v}{b}$

so final answer will be $L = \frac{mvb}{6}$

7 0

2 years ago