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4 years ago

Two things you can do to increase the acceleration of an object

Physics

1 answer:

stiv31 [10]4 years ago

7 0

You can decrease the mass, or you can increase the force applied to the object

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A rubber ball and a lump of clay have equal mass. They are thrown with equal speed against a wall. The ball bounces back with ne

gregori [183]

Answer:

The ball experiences the greater momentum change

Explanation:

The momentum change of each object is given by:

$\Delta p = m \Delta v= m (v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

Both objects have same mass m and same initial velocity u. So we have:

- For the ball, the final velocity is

$v=-u$

Since it bounces back (so, opposite direction --> negative sign) with same speed (so, the magnitude of the final velocity is still u). So the change in momentum is

$\Delta p=m(v-u)=m((-u)-u)=-2mu$

- For the clay, the final velocity is

$v=0$

since it sticks to the wall. So, the change in momentum is

$\Delta p = m(v-u)=m(0-u)=-mu$

So we see that the greater momentum change (in magnitude) is experienced by the ball.

3 0

3 years ago

a. 2.00kg object is subject to three force that gives acceleration a =8m/s^2 i +6m/s j if two of three forces are f1 =(30.0N)+(1

tekilochka [14]

By Newton's second law, the net force on the object is

∑ F = m a

∑ F = (2.00 kg) (8 i + 6 j ) m/s^2 = (16.0 i + 12.0 j ) N

Let f be the unknown force. Then

∑ F = (30.0 i + 16 j ) N + (-12.0 i + 8.0 j ) N + f

=> f = (-2.0 i - 12.0 j ) N

8 0

3 years ago

A complete path through which charge can flow is an electric what?

grigory [225]

A complete path through which charge can flow is an "Electric Circuit"

Hope this helps!

5 0

3 years ago

Read 2 more answers

A flap of tissue that prevents blood from flowing back is a(n)________ .

ioda

A valve is a flap of tissue that prevents blood from flowing back

6 0

3 years ago

A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3

4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

$B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT$

8 0

3 years ago