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3 years ago

A 1000 kg car is rolling slowly across a level surface at 1m/s, heading toward a group of small innocent children.

Physics

2 answers:

Ne4ueva [31]3 years ago

7 0

Answer:

The force is -1620.73 N.

Explanation:

Given that,

Mass of car = 1000 kg

Velocity = 1 m/s

Distance = 2 m

Angle = 30°

We need to calculate the force

Using formula of work done

$W_{net}=K_{f}-K_{i}$

$F\times d=K_{f}-K_{i}$

$Fd\cos\theta=-K_{i}$

$F=\dfrac{-K_{f}}{d\cos\theta}$

$F=-\dfrac{\dfrac{1}{2}mv^2}{d\cos\theta}$

Put the value into the formula

$F=-\dfrac{\dfrac{1}{2}\times1000\times(1)^2}{2\times\cos30}$

$F=-1620.73\ N$

Hence, The force is -1620.73 N.

AnnyKZ [126]3 years ago

4 0

Answer:

Force, F = 288.67 N

Explanation:

Mass of the car, m = 1000 kg

Initial speed of the car, u = 1 m/s

The car and push on the hood at an angle of 30° below horizontal, $\theta=30^{\circ}$

Distance, d = 2 m

Let F is the force must you push to stop the car. using work energy theorem to find it. According to this theorem, the work done is equal to the change in kinetic energy as :

$W=\dfrac{1}{2}m(v^2-u^2)$

$F\times d=\dfrac{1}{2}m(v^2-u^2)$

It stops, v = 0

$Fd\ cos\theta=\dfrac{1}{2}m(u^2)$

$F=\dfrac{\dfrac{1}{2}m(u^2)}{d\ cos\theta}$

$F=\dfrac{\dfrac{1}{2}\times 1000\times (1)^2}{2\ cos(30)}$

F = -288.67 N

So, the force required to push to stop the car is 288.67 N. Hence, this is the required solution.

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When you double the speed of your car, your braking distance quadruples. every time you double your speed, you multiply your braking distance by four.

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2 years ago

If a car has an initial velocity of 20 m/s and accelerates at 2.0 m/s2 for 100 m, what is its final velocity?

rodikova [14]

Answer:

28.28 ms-1

Explanation:

v² = u² + 2as

v² = 20² + 2×2×100

v²= 800

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3 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

2 years ago

Describe the general relationship between the weight of adult men and the average number of Calories needed daily to maintain a

Alexeev081 [22]

Answer:

The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking.

The amount of calories contained in the food we eat every day must represent the amount of calories eliminated by the body in that time to have a steady weight.

Explanation:

The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking. If total quantity of calories in the food we consume every day is higher that the average number of calories needed daily, then weight increases by fat accumulation.

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2 years ago

The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,

Travka [436]

1) In a circular motion, the angular displacement $\theta$ is given by
$\theta = \frac{S}{r}$
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: $S=2600 m$ . Using the information about the radius, $r=0.35 m$ , we find the total angular displacement:
$\theta = \frac{2600 m}{0.35 m} =7428 rad$

2) If we put larger tires, with radius $r=0.60 m$ , the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
$\theta = \frac{2600 m}{0.60 m}=4333 rad$

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3 years ago