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zheka24 [161]
2 years ago
6

Find the displacement the body in the following graph​

Physics
1 answer:
Len [333]2 years ago
7 0

Answer:

150

Explanation:

15m/s×10s= 15m/a

and that is the correct answer

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How strong is the electric field between two parallel plates 4.2 mm apart if the potential difference between them is 220 V?
NISA [10]

Answer:

2200000 \div 42

6 0
3 years ago
After a day of testing race cars, you decide to take your own 1550 kg car onto the test track. While moving down the track at 10
quester [9]

Answer:

F = 2325 N

Explanation:

given,

mass of the car, m = 1550 Kg

initial speed, u = 10 m/s

final speed, v = 25 m/s

time, t = 10 s

Average net force = ?

acceleration of the car

a = \dfrac{v-u}{t}

a = \dfrac{25-10}{10}

a = 1.5 m/s²

we know

F = m a

F = 1550 x 1.5

F = 2325 N

Net average force applied on the car is equal to 2325 N.

6 0
3 years ago
Read 2 more answers
Two balls with equal masses, m, and equal speed, v, engage in a head on elastic collision. what is the final velocity of each ba
Allushta [10]
The collision is elastic. This means that both momentum and kinetic energy are conserved after the collision.

- Let's start with conservation of momentum. The initial momentum of the total system is the sum of the momenta of the two balls, but we should put a negative sign in front of the velocity of the second ball, because it travels in the opposite direction of ball 1. So ball 1 has mass m and speed v, while ball 2 has mass m and speed -v:
p_i = p_1-p_2 = mv-mv =0
So, the final momentum must be zero as well:
p_f = 0
Calling v1 and v2 the velocities of the two balls after the collision, the final momentum can be written as
p_f = mv_1 + mv_2 = 0
From which
v_1 = -v_2

- So now let's apply conservation of kinetic energy. The kinetic energy of each ball is \frac{1}{2} mv^2. Therefore, the total kinetic energy before the collision is
K_i = \frac{1}{2} mv^2 +  \frac{1}{2} mv^2 = mv^2
the kinetic energy after the collision must be conserved, and therefore must be equal to this value:
K_f = K_i = mv^2 (1)
But the final kinetic energy, Kf, is also
K_f =  \frac{1}{2} mv_1^2 +  \frac{1}{2}mv_2^2
Substituting v_1 = -v_2 as we found in the conservation of momentum, this becomes
K_f = mv_2 ^2
we also said that Kf must be equal to the initial kinetic energy (1), therefore we can write 
mv_2^2 = mv^2

Therefore, the two final speeds of the balls are
v_2 = v
v_1 = -v_2 = -v

This means that after the collision, the two balls have same velocity v, but they go in the opposite direction with respect to their original direction.

8 0
3 years ago
Go to the Previous QuestionGo to the Next Question
Liono4ka [1.6K]

Answer:

hard2

Explanation:

4 0
2 years ago
1. A 2-kg bowling ball sits on top of a building that is 40 meters tall.
defon

Answer: GPE = 784 J, K.E. = 0

Explanation:

An object possess gravitational potential energy due to virtue of its position and kinetic energy due to virtue of its motion. A bowling possesses only gravitational potential energy due to its position. Its not in motion, thus its kinetic energy is zero.

Gravitational potential energy is given by:

G.P.E = m g h

where m is the mass, g is acceleration due to gravity and h is the height.

m = 2.0 kg

g = 9.8 m/s²

h = 40.0 m

G.P.E = 2.0 kg × 9.8 m/s² × 40.0 m = 784.0 J

K.E. = 0.5 m v²

∵ v = 0 ⇒ K.E. = 0

8 0
3 years ago
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