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kolbaska11 [484]

3 years ago

15

0.A 20-g bullet moving at 1 000 m/s is fired through a one-kg block of wood emerging at a speed of 100 m/s. What is the kinetic

energy of the block that results from the collision if the block had not been moving prior to the collision and was free to move?

1 answer:

Fiesta28 [93]3 years ago

4 0

Answer:

KE = 0.162 KJ

Explanation:

given,

mass of bullet (m)= 20 g = 0.02 Kg

speed of the bullet (u)= 1000 m/s

mass of block(M) = 1 Kg

velocity of bullet after collision (v)= 100 m/s

kinetic energy = ?

using conservation of momentum

m u = m v + M V

0.02 x 1000 = 0.02 x 100 + 1 x V

20 = 2 + V

V = 18 m/s

now,

Kinetic energy of the block

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

$KE = \dfrac{1}{2}\times 1 \times 18^2$

KE = 162 J

KE = 0.162 KJ

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A nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the

Sladkaya [172]

Answer:

(a). The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

$V = 0$

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

$V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr$

$V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr$

$V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}$

$V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})$

$V_{r}=-0.00012190\ V$

$V_{r}=-1.219\times10^{-4}\ V$

The electric potential at 1.650 cm is $-1.219\times10^{-4}\ V$ .

(b). We need to calculate the potential at a distance r = R

Using formula of potential difference

$V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}$

$V_{R}=-0.0003759\ V$

$V_{R}=-3.759\times10^{-4}\ V$

The electric potential at 2.81 cm is $-3.759\times10^{-4}\ V$ .

Hence, This is the required solution.

5 0

3 years ago

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2551

RUDIKE [14]

Answer:

0.7 kg m²

Explanation:

F = force exerted applied by muscle in a professional boxer = 2551 N

r = length of lever arm = 3.15 cm = 0.0315 m

α = angular acceleration of the forearm = 115 rad/s²

I = moment of inertia of the boxer's forearm

τ = Torque applied by muscle in a professional boxer

Torque is given as

τ = I α = r F

Inserting the values

I (115) = (0.0315) (2551)

I = 0.7 kg m²

7 0

3 years ago

A neighbor's child wants to go to a neighborhood carnival to experience the wild rides. The neighbor is worried about safety as

seraphim [82]

Answer:

No, the ride is not safe

Explanation:

From the diagram attached, it is seen that

$\sum f_{y} = 0$

$\tau cos \theta - W = 0$ ................(1)

$\sum f_{x} = 0$

$-ma + \tau sin \theta = 0$ .......(2)

$a = w^{2} r$

From the diagram, $r = Lsin \theta$

$a = w^{2} L sin \theta$

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

$a = \frac{4\pi ^{2}L sin \theta }{T^{2} }$ ............(3)

Put (3) into (2)

$\tau sin \theta = \frac{4\pi ^{2}m L sin \theta }{T^{2} }$

$\tau = \frac{4\pi ^{2}m L }{T^{2} }$

w = weight of chair + weight of child = 50 + 10 = 60 lb

g = 32 ft/s²

m = w/g = 60/32 = 1.875

L = 30 ft

T = 3 secs

$\tau = \frac{4\pi ^{2}*1.875*30 }{0.3^{2} }$

$\tau = 246.74 lbs$

since 246.74 lbs > 200 lbs, it is not safe because the stationary chair will creak

6 0

3 years ago

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When heat is added to an object, what is happening to the item at the atomic level? (Check all that apply)

3241004551 [841]

Answer:

a

Explanation:

heat is energy, energy cannot be made or destroyed but transferred

6 0

3 years ago

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Initially, a particle is moving at 5.25 m/s at an angle of 35.5Â° above the horizontal. Three seconds later, its velocity is 6.0

ivolga24 [154]

Answer:

a =( -0.32 i ^ - 2,697 j ^) m/s²

Explanation:

This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.

Break down the speeds in two moments

initial

v₀ₓ = v₀ cos θ

v₀ₓ = 5.25 cos 35.5

v₀ₓ = 4.27 m / s

$v_{oy}$ = v₀ sin θ

$v_{oy}$ = 5.25 sin35.5

$v_{oy}$ = 3.05 m / s

Final

vₓ = 6.03 cos (-56.7)

vₓ = 3.31 m / s

$v_{y}$ = v₀ sin θ

$v_{y}$ = 6.03 sin (-56.7)

$v_{y}$ = -5.04 m / s

Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order

a = ( $v_{f}$ - v₀) /t

aₓ = (3.31 -4.27)/3

aₓ = -0.32 m/s²

$a_{y}$ = (-5.04-3.05)/3

$a_{y}$ = -2.697 m/s²

6 0

3 years ago