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Pepsi [2]
3 years ago
10

Powerful sports cars can go from zero to 25 m/s (about 60 mph) in 6 seconds. What is the magnitude of the acceleration, includin

g units
Physics
1 answer:
Vikentia [17]3 years ago
7 0

Answer:

The magnitude of the acceleration is 4.2 m/s²

Explanation:

Given;

initial velocity of the powerful sports car, u = 0

final velocity of the powerful sports car, v = 25 m/s

time of motion, t = 6 seconds

Acceleration = Δv / Δt

Δv is change in velocity

Δt is change in time

The magnitude of the acceleration of the powerful sports car during the motion is given by;

a = \frac{v-u}{t}\\\\a = \frac{25-0}{6}\\\\a = 4.2 \ m/s^2

Therefore, the magnitude of the acceleration is 4.2 m/s²

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Does the lattice energy of an ionic solid increase or decrease as the charges or sizes of the ions increase?
alexandr1967 [171]

Answer:

Explanation:

Lattice energy is the energy required to separate one mole of an ionic solid compound into its components gaseous cations and anions.

Due to increase in size of the ions, the lattice energy decreases while the lattice energy increases as the charge of the ions increases.

When the size increase, the distance between the nuclei also increase leading a decrease the force of attraction between the nuclei

7 0
3 years ago
1. If a car travels 400m in 20 seconds how fast is it going?
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Answer:

1) 20 m/s

2) 5 m/s

3) 2 m/s

4) 395,000m/9000s

5) 16 km/0.25h

Explanation:

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3 years ago
What is the distance from one crest to the next crest?
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3 0
3 years ago
Can someone help me please
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D

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8 0
3 years ago
Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col
Ratling [72]
In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
p_f=(m+M)v_f=4.8 kg m/s
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find v_f, the velocity of the nail and the block after the collision:
v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}=  0.48 m/s

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
p_i = mv_i = p_f
Rearranging the formula, we can find v_i, the velocity of the nail before the collision:
v_i =  \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s
6 0
3 years ago
Read 2 more answers
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