Calculate the kinetic energy of a 1500 kg car moving at 42 km/hr.

Vlad1618 [11]

B: heat is transferred as thermal energy by the interaction of moving particles

6 0

3 years ago

The small piston of a hydraulic lift has a diameter of 8.0 cm, and its large piston has a diameter of 40cm. the lift raises a lo

Elena L [17]

Sm = small piston
la = large piston

P=F/A
P=15000/(20^2)π

F of sm = PA
= (75/2π)•((8^2)π)
= (75•64)/2
= 4800/2
= 2400N
We already know the pressure but giving it in an approximate decimal form, to two significant figures (since that's what your supplied precision is at):

a) 12 Pa
b) 2400 N

7 0

3 years ago

Read 2 more answers

Maverick and goose are flying a training mission in their F-14. They are flying at an altitude of 1500 m and are traveling at 68

lions [1.4K]

Answer:

The bomb will remain in air for <u>17.5 s</u> before hitting the ground.

Explanation:

Given:

Initial vertical height is, $y_0=1500\ m$

Initial horizontal velocity is, $u_x=688\ m/s$

Initial vertical velocity is, $u_y=0(\textrm{Horizontal velocity only initially)}$

Let the time taken by the bomb to reach the ground be 't'.

So, consider the equation of motion of the bomb in the vertical direction.

The displacement of the bomb vertically is $S=y-y_0=0-1500=-1500\ m$

Acceleration in the vertical direction is due to gravity, $g=-9.8\ m/s^2$

Therefore, the displacement of the bomb is given as:

$S=u_yt+\frac{1}{2}gt^2\\-1500=0-\frac{1}{2}(9.8)(t^2)\\1500=4.9t^2\\t^2=\frac{1500}{4.9}\\t=\sqrt{\frac{1500}{4.9}}=17.5\ s$

So, the bomb will remain in air for 17.5 s before hitting the ground.

5 0

3 years ago

A chunk of metal has a volume of

Alla [95]

Answer:

7213.7kg/m³

Explanation:

Given parameters:

Volume of the metal chunk = 0.131m³

Mass of the metal chunk = 945kg

Unknown:

Density of the metal chunk = ?

Solution :

To solve this problem, density is the mass per unit volume. It is mathematically expressed as;

Density = $\frac{mass}{volume}$

So;

Density = $\frac{945}{0.131}$ = 7213.7kg/m³

5 0

3 years ago

A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static fr

PSYCHO15rus [73]

Answer:

ω = 2.1 rad/sec

Explanation:

As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
Now, we need to ask ourselves: what supplies this force?
In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
We know that this force can be expressed as follows:

$f_{frs} = \mu_{s} * F_{n} (1)$

where μs = coefficient of static friction between the rock and the merry-

go-round surface = 0.7, and Fn = normal force.

In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
Fn = m*g (2)
This static friction force is just the same as the centripetal force.
The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

$F_{c} = m* \omega^{2}*r (3)$

Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

$\omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec$

This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.

4 0

3 years ago