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allsm [11]
3 years ago
11

2 objects have a total momentum of 400kg m/s, they collide. Object A’s mass is5kg & object B’s mass is 11kg. After the colli

sion Object B is moving at 15m/s.What is the velocity of Object A AFTER the collision?
Physics
1 answer:
ss7ja [257]3 years ago
3 0

Answer:

Explanation:

We shall apply law of conservation of momentum .

Momentum before collision = momentum after collision .

Momentum before collision = 400 kg m/s

Momentum after collision = 5  x v + 11 x 15

where v is velocity of A after the collision .

5  x v + 11 x 15 = 400

5 v = 400 - 165

5v = 235

v = 47 m /s .

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A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 100 N is being applied to the sled rope
tangare [24]

Answer:

60.18 N

Explanation:

Given that:

The force applied on the sled = 100 N

Suppose, the angle between the sled rope and the ground = 53°

The horizontal force which acts  in the horizontal direction can be expressed as:

F_x = F \ cos \theta

F_x = 100 \ cos (53)

F_x = 60.18 \ N

But if the angle between the sled rope is parallel to the ground. Then, we use an angle on a straight line which is = 180°

F_x = F \ cos \theta

F_x = 100 \ cos (180)

= 100 × -1

= -100 N

3 0
3 years ago
A 95 kg fullback, running at 8.2 m/s, collides in midair with a 128 kg defensive tackle moving in the opposite direction. Both p
Sindrei [870]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

Below is the answers:

Fullback running 

<span>Mo = mass * velocity </span>
<span>Mo = 95kg * 8.2 m/s =779 kg*m/s (a </span>

<span>He got stopped Change in Mo = 779 kg*m/s (b </span>

<span>Both stopped ===> Tackle's mo = - Halfback's Mo = - 779 kg*m/s (c & d </span>

<span>- 779 = 128 * v </span>
<span>v= - 6.09 m/s (e</span>
6 0
3 years ago
Two conducting spheres of different sizes are at the same potential. The radius of the larger sphere is four times larger than t
Sergeeva-Olga [200]

Answer:

0.8

Explanation:

The two spheres have the same potential, V.

Let the radius of the larger sphere be R and the radius of the smaller sphere be r,

=> R = 4r

Let the charge on the smaller sphere be q. Hence, the larger sphere will have charge Q - q.

The potential of the smaller sphere will be:

V_S = \frac{kq}{r}

The potential of the larger sphere will be:

V_L = \frac{k(Q - q)}{R}

Inputting R = 4r,

V_L = \frac{k(Q - q)}{4r}

Since V_S = V_L = V,

\frac{k(Q - q)}{4r} = \frac{kq}{r}

=> Q - q = 4q

=> 5q = Q

q = 0.2Q

The fraction of the charge Q that rests on the smaller sphere is 0.2

The charge of the larger sphere is:

Q - q = Q - 0.2Q = 0.8Q

∴ The fraction of the total charge Q that rests on the larger sphere is 0.8

7 0
3 years ago
Which of the following is NOT a function of MITOSIS?
DIA [1.3K]

Answer:

sexual reproduction

Explanation:

while sexual reproduction is a function of meiosis

6 0
3 years ago
True or false? Nitrogen in the atmosphere is responsible for clouds and precipitation? If not what is?
Flura [38]
A.)

False
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</u><u>T</u>here reason why it's false is because Nitrogen is not responsible for clouds and precipitation. The real answer is that Water (Ocean / Other) is responsible for clouds and precipitation. 

Good Day / Night :D

6 0
3 years ago
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