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faltersainse [42]
3 years ago
12

A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s.

Physics
2 answers:
ELEN [110]3 years ago
8 0
Momentum = mass x velocity

Before collision
Momentum 1 = 2 kg x 20 m /s = 40 kg x m/s
Momentum 2 = 3 kg x -10m/s = -30 kg x m/s

After collision
Momentum 1 = 2 kg x -5 m/s = -10 m/s
Momentum 2 = 3 kg x V2 = 3V2

Total momentum before = total momentum after
40 + -30 = -10 + 3V2
V2 = <span>6.67 m/s

Total kinetic energy before
</span><span>= (1/2) [ 2 kg * 20 m/s * 2 + 3 kg * ( -10 m/s) *2 ]
= 550 J
</span>
<span>Total kinetic energy after
</span>= (1/2) [ 2 kg * ( - 5 m/s) * 2 + 3 kg * 6.67 m/s *2 ]
= 91.73 J

Total kinetic energy lost during collision
=<span>550 J - 91.73 J
= 458.27 J</span>

Stells [14]3 years ago
4 0
Kinetic energy before:

E = 0.5mv^2 + 0.5mv^2

= 0.5*2*20^2 + 0.5*3*(-10)^2

= 400 + 150

= 550J

Kinetic energy after:

E = 0.5mv^2

= 0.5*2*(-5)^2

= 25J

Kinetic energy lost during the collision:

E = 550 - 25

= 525J
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Which is not an advantage of radio telescopes over optical telescopes?
amid [387]
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8 0
3 years ago
Two upwards forces act: one of 34N which is 3.5m to the left of the axle, the other is 25N and is 2.6m to the right of the axle.
KiRa [710]

Answer:

See explanation below

Explanation:

In this case, you want to know if you put an object between these forces, which direction would go.

To know this, we need to calculate the moment of an object, which is defined as the product of a force and it's distance. In other words:

M = F * d   (1)

And, in order to reach equilibrium the force will exert a direction in clockwise or anticlosewise, and these moments, should be even:

anticlockwise moment = clockwise moment.

The clockwise would be the forces to the right, and anticlock would the only force to the left of the axle.

Clockwise moment = (10 * 0.8) + (25 * 2.6) = 73 Ns

Anticlockwise moment = 34 * 3.5 = 119 Ns.

As we can see, the moment in the anticlockwise is higher than the actual clockwise moment, therefore, we can assume that the object will move anticlockwise, or simply move to the left.

Hope this helps

5 0
3 years ago
Products must equal which in a chemical reaction?
Evgesh-ka [11]

I think it is reactants

3 0
3 years ago
A rock is thrown from the top of a 20-m building at an angle of 53° above the horizontal. If the horizontal range of the throw
NemiM [27]

Answer:

28.5 m/s

18.22 m/s

Explanation:

h = 20 m, R = 20 m, theta = 53 degree

Let the speed of throwing is u and the speed with which it strikes the ground is v.

Horizontal distance, R = horizontal velocity x time

Let t be the time taken

20 = u Cos 53 x t

u t = 20/0.6 = 33.33 ..... (1)

Now use second equation of motion in vertical direction

h = u Sin 53 t - 1/2 g t^2

20 = 33.33 x 0.8 - 4.9 t^2     (ut = 33.33 from equation 1)

t = 1.17 s

Put in equation (1)

u = 33.33 / 1.17 = 28.5 m/s

Let v be the velocity just before striking the ground

vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s

vy = uSin 53 - 9.8 x 1.17

vy = 28.5 x 0.8 - 16.66

vy = 6.14 m/s

v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2

v = 18.22 m/s

6 0
3 years ago
The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont
ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the  object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the  object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}

But heat capacity of object B is twice that of object A

T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K

Therefore, the final temperature of both objects is 400 K.

4 0
3 years ago
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