A 10.0 cm3 sample of copper has a mass of 89.6

An object of mass m moves horizontally, increasing in speed from 0 to v in a time t. The power necessary to accelerate the objec

lana66690 [7]

Answer:

Option e is correct.

Explanation:

An object of mass m moves horizontally, increasing in speed from 0 to v in a time t.

We know that,

Work done = change in kinetic energy

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

Here, u = 0

So,

$W=\dfrac{1}{2}mv^2$

Also,

Power = work done/time

So,

$P=\dfrac{\dfrac{1}{2}mv^2}{t}\\\\=\dfrac{mv^2}{2t}$

So, the correct option is (e).

8 0

2 years ago

If you were trying to cross a river with the shortest possible time, would you aim your boat slightly upstream, directly across

Maru [420]

Slightly downstream for the shortest possible time

7 0

2 years ago

A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache

Paladinen [302]

Answer:

$F_a=1470\ N$

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

$\displaystyle F_a-F_{r1}-F_{r2}=m.a=0$

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

$\displaystyle F_a=F_{r1}+F_{r2}.....[1]$

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

$\displaystyle F_{r2}-T=0$

The friction forces are computed by

$\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g$

$\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g$

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

$\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g$

Simplifying

$\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)$

Plugging in the values

$\displaystyle F_{a}=0.25(9.8)[400+2(100)]$

$\boxed{F_a=1470\ N}$

8 0

2 years ago

A gymnast dismounts the uneven parallel bars with some angular momentum about her transverse axis. Just after release, she is in

Luda [366]

Answer:

a. Her moment of inertia increases and she rotates slower.

Explanation:

As we know that initially when she starts her motion she is in piked position due to which her whole mass is concentrated near the axis of rotation

So here the rotational Inertia of her body will be smaller

Now when is comes closer to the position of landing she extends into layout position due to which her mass will move away from the axis of rotation

Due to this the rotational inertia of her body will increase

now we know that there is no external torque on the system

so here angular momentum must be conserved

So we will have

$I\omega = constant$

so if rotational inertia is increasing then angular speed must be slower

so correct answer will be

a. Her moment of inertia increases and she rotates slower.

7 0

2 years ago

The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm

Nesterboy [21]

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

8 0

2 years ago