The final speed of the car is 2) 150 m/s
Explanation:
Since the motion of the car is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the car in this problem, we have
u = 10 m/s

s = 7,467 m
Solving for v, we find the final velocity (and speed) of the car:

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Answer:
The time taken for the race is 17.20 s.
Explanation:
It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.
Calculate the final speed of the sprinter.
The expression for the equation of the motion is as follows;

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.
Put u= 0, s=30 m and
.


Calculate time taken to cover 30 m distance.
The expression for the equation of motion is as follows;

Put u= 0, s=30 m and
.

t=6.45 s
Calculate the time taken to complete his race.
T= t+t'
Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

Put s= 30 m,
and s'= 100 m.

T= 17.20 s
Therefore, the time taken for the race is 17.20 s.
Answer:
Nucleus A with 7 protons and 7 neutrons and Nucleus C with 7 protons and 5 neutrons are isotopes of the same elements
Explanation:
Isotopes are elements that have the same atomic structure but different molecular structure. An atom that has the same atomic number but different mass number are known to be isotopes.
The proton of an atom is the same as its atomic number while the sum of the proton and neutron is equal to its mass number.
According to the question, nuclei that has the same number of proton are isotopes of the same element. Therefore nuclei A and C with 7 protons each are isotopes of the same element since they have the same atomic number i.e number of proton = atomic number.
Their atomic masses of nuclei A and C are 14 and 12 respectively
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Answer:
3.75 m/s
Explanation:
From the question given above, the following data were obtained:
Acceleration (a) = 1.5 m/s²
Initial velocity (u) = 0 m/s
Time (t) = 2.5 s
Final velocity (v) =?
a = (v – u) / t
1.5 = (v – 0) / 2.5
1.5 = v / 2.5
Cross multiply
v = 1.5 × 2.5
v = 3.75 m/s
Hence, the escape velocity of the squirrel is 3.75 m/s