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3 years ago

A car possesses 20 000 units of momentum. What would be the car's new momentum if its mass was doubled

Physics

1 answer:

Klio2033 [76]3 years ago

7 0

Answer:

40 000 units of momentum

Explanation:

Assuming that the mass is doubled and the velocity stays constant the new momentum will be doubled the original momentum.

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A jewellery shop owner has two identical clear gemstones but cannot remember which one is diamond and which one is rutile. Rutil

pantera1 [17]

We know

$\boxed{\sf n_{21}=\dfrac{C}{V}}$

How he find:-

The owner will measure speed of light through the index(Diamond or rutile).Then using calculations he fill find which is the velocity of diamond or rutile

<h3> For diamond</h3>

$\\ \sf\longmapsto n_D=\dfrac{C}{V_D}$

$\\ \sf\longmapsto 2.4=\dfrac{3\times 10^8ms^{-1}}{V_D}$

$\\ \sf\longmapsto V_D=\dfrac{3\times 10^8ms^{-1}}{2.4}$

$\\ \sf\longmapsto V_D=1.25\times 10^8ms^{-1}$

$\\ \sf\longmapsto V_D=125\times 10^6ms^{-1}$

<h3>For rutile</h3>

$\\ \sf\longmapsto n_R=\dfrac{C}{V_R}$

$\\ \sf\longmapsto V_R=\dfrac{C}{n_R}$

$\\ \sf\longmapsto V_R=\dfrac{3\times 10^8ms^{-1}}{2.9}$

$\\ \sf\longmapsto V_R=1.03\times 10^8$

$\\ \sf\longmapsto V_R=103\times 10^6ms^{-1}$

5 0

3 years ago

You expend 1000 w of power moving a piano 10 meters in 5 seconds. how much force do you exert

Vinvika [58]

Answer:

Force, F = 500 N

Explanation:

Given that,

Power expanded, P = 1000 W

Distance moved by piano, d = 10 m

Time taken, t = 5 s

To find,

The force exerted by the person

Solution,

The rate at which the work is being done is called power delivered. Its formula is given by:

$P=\dfrac{W}{t}$

W is the work done, W = F × d

$P=\dfrac{Fd}{t}$

On rearranging the above formula to find F as :

$F=\dfrac{P t}{d}$

$F=\dfrac{1000\times 5}{10}$

F = 500 N

Therefore, the force exerted on the person is 500 N. Hence, this is the required solution.

7 0

3 years ago

A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (

Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants ( $v_{oy}$ = 0) zero, so let's use the equation

y = $v_{oy}$ t -1/2 g t²

y= - ½ g t²

t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

x = vox t

x = v₀ₓ √2y/g

c) Speeds before touching the ground

vₓ = vox = constant

$v_{y}$ = $v_{oy}$ - gt

$v_{y}$ = 0 - g √2y/g

$v_{y}$ = - √2gy

tan θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0

4 years ago

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn

Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

$F=mrw^2$

$m=$ mass of the corresponding bugs

$r=$ corresponding radial distance of each bug

$w=$ angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

$F$ ∝ $mr$

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0

3 years ago

Read 2 more answers

A force of 6.0 N is applied horizontally to a 3.0 kg crate initially at rest on a horizontal frictionless surface. After the cra

Savatey [412]

Answer:

a. Yes, because the acceleration of the crate is 2.0 m/s².

Explanation:

Given

$Force = 6N$ --- f