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2 years ago

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A 1500 kg car is pushing a 4000 kg truck. The car and truck are accelerating at 2.0 m/s^2. Assuming that the frictional force on

the truck is negligible, what force is the truck exerting on the car?

1 answer:

Crazy boy [7]2 years ago

7 0

To solve this problem we will use the Force equation according to the definition given in Newton's second law. There we have that the Force is equal to

$F =ma$

Where,

m = mass

a = acceleration

Our values are given as

$m_1 = 1500 kg$

$m_2 = 4000 kg$

$a = 2.0 m/s^2$

Considering that both mass are equal to one, we have that:

$F = (m_1+m_2)* a$

$F = (1500 + 4000)(2.0)$

$F = 11000 N = 11kN$

Therefore the truck exert a force on the car of 11kN

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According to the nebular theory of solar system formation, what key difference in their early formation explains why the jovian

svp [43]

Answer:

The Jovian planets formed beyond the Frostline while the terrestrial planets formed in the Frostline in the solar nebular

Explanation:

The Jovian planets are the large planets namely Saturn, Jupiter, Uranus, and Neptune. The terrestrial planets include the Earth, Mercury, Mars, and Venus. According to the nebular theory of solar system formation, the terrestrial planets were formed from silicates and metals. They also had high boiling points which made it possible for them to be located very close to the sun.

The Jovian planets formed beyond the Frostline. This is an area that can support the planets that were made up of icy elements. The large size of the Jovian planets is as a result of the fact that the icy elements were more in number than the metal components of the terrestrial planets.

3 0

2 years ago

Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o

sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

$a_{x} = -\omega^{2} x$

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders is given as:

$I=\frac{1}{2}MR^{2}$ -----(1)

Newton's second law for angular motion is:

∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

∑Fₓ = Maₓ ------ (3)

By definition of torque:

τ = RF --------(4)

Put (4) and (1) in (2)

$RF=\frac{1}{2}MR^{2}\alpha$

$RF=\frac{1}{2}MR^{2}\alpha$

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

$f_{s}=\frac{1}{2}MR\alpha$ ------ (5)

As angular acceleration is related to linear by:

$a= R\alpha$

Eq (5) becomes

$f_{s}=\frac{1}{2}Ma_{x}$ ---- (6)

aₓ shows displacement in horizontal direction

From (3)

∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

$\sum F_{x} = f_{s} - kx$ ----(7)

Put (7) in (3)

$f_{s} - kx = Ma_{x}$ [/tex] -----(8)

Using (6) in (8)

$\frac{1}{2}Ma_{x} - kx =Ma_{x}$

$a_{x} = \frac{2k}{3M} x$ --- (9)

For spring mass system

$a= -\omega^{2} x$ ----- (10)

Equating (9) and (10)

$\omega^{2} = \frac{2k}{3M}$

$\omega = \sqrt{ \frac{2k}{3M}}$

then (9) becomes

$a_{x} = - \omega^{2}x$

(The minus sign says that x and aₓ have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

$T=\frac{2\pi }{\omega}$

$T = 2\pi \sqrt{\frac{3M}{2k}$

5 0

2 years ago

A cyclist moving with a constant velocity of 6.0 m/s forward passes a car that is just starting. If the car has a constant accel

sergiy2304 [10]

After 6 seconds, the car will surpass the cyclist.

<h3><u>Explanation:</u></h3>

The speed of the cyclist = 6 m/s.

Let after time t sec, the car will overtake the cyclist.

So, distance covered by the cyclist in t sec = 6t m

Initial velocity of the car is 0 m/s, because the car is just starting.

Acceleration of the car = $2 m/s^2$ .

Final velocity of the car =6 m/s.

So to cover the distance 6t, the time required by the car = $\frac{1}{2} \times a \times t^2 = \frac{1}{2} \times 2 \times t^2$

$6t = \frac{1}{2} \times 2 \times t^2\\6t = t^2$

t =6 sec

So, after 6 seconds, the car will surpass the cycle.

6 0

2 years ago

A tennis ball of mass m=0.060 kg and speed v=25 m/s strikes a wall at a 45 angle and rebounds with the same velocity at 45°. Wha

Diano4ka-milaya [45]

To solve this problem we will apply the concepts related to the Impulse which can be defined as the product between mass and the total change in velocity. That is to say

$p = m\Delta v$

Here,

m = mass

$\Delta v =$ Change in velocity

As we can see there are two types of velocity at the moment the object makes the impact,

the first would be the initial velocity perpendicular to the wall and the final velocity perpendicular to the wall.

That is to say,

$v_i = vcos\theta$

$v_f = -v sin\theta$

El angulo dado es de 45° y la velocidad de 25, por tanto

$v_i = (25)cos(45) = 17.68m/s$

$v_f = -(25)sin(45) = -17.68m/s$

The change of sign indicates a change in the direction of the object.

Therefore the impulse would be as

$p = 0.060(-17.68-17.68)$

$p = -2.12kg \cdot m/s$

The negative sign indicates that the pulse is in the opposite direction of the initial velocity.

3 0

2 years ago

Water evaporates off lakes. Winds blow across the planet. Where does the energy come from for these and other weather processes?

Otrada [13]

Answer:

B. Solar energy

Explanation:

The water cycle is driven primarily by the energy from the sun. This solar energy drives the cycle by evaporating water from the oceans, lakes, rivers, and even the soil. Other water moves from plants to the atmosphere through the process of transpiration.

8 0

2 years ago

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