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2 years ago

PLEASE HELP MEE PLEASE I BEG

Physics

1 answer:

TEA [102]2 years ago

8 0

Explanation: (I think)

Plug your values into the momentum equation.

So m1= 63kg

m2 = 10 kg

V1 = 12 m/s

And then plug in your values and solve for your unknown (v2)

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39. What is the change in momentum for a 5,000 kg ship in

Ivan

Answer:

Change in momentum is zero.

Explanation:

The following data were obtained from the question:

Mass (m) = 5000 kg

Time (t) = 1 h

Net force (F) = 0

Change in momentum =?

Force = Rate of change of momentum

0 = change in momentum

Change in momentum = 0

We can see from the above illustration that the net force is zero. Thus, the change in momentum is also zero.

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2 years ago

What the opposite of suspension physical science physical science

Svetlanka [38]

A colloid I think. Don’t hold it against me if I’m wrong my dude.

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2 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring

azamat

Answer:

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120 $\frac{N}{m}$

Angle $\theta$ = 34°

From the free body diagram

Force acting on the box = mg sin $\theta$

⇒ F = 3 × 9.81 × $\sin34$

⇒ F = 16.45 N ------- (1)

Since box is attached with the spring so a spring force also acts on the box.

$F_{sp}$ = k x

$F_{sp}$ = 120 $x$ -------- (2)

The net force acting on the body is given by

$F_{net} = ma$

Since acceleration of the box is zero so

$F_{net} = 0$

$F - F_{sp} = 0$

$F = F_{sp}$

Put the values from equation (1) & (2) we get

16.45 = 120 $x$

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

3 0

3 years ago

Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before stri

mixas84 [53]

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o$ = 5.50 m/s

Her speed after entering the water, $V_f$ = 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times dT$

Here, F = the force ,

dT = time interval over which the force is applied for

= 1.65 s

dP = change in momentum

dP = m x dV

$= m \times [V_f - V_o]$

= 82 x (1.1 - 5.5)

= -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

= 218 N

8 0

2 years ago