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Vlad1618 [11]
2 years ago
6

A racecar driver on a flat racetrack steps on the gas, changing her velocity from 10 m/sec to 30 m/sec in 5 seconds. What is the

acceleration of the racecar during that time frame?
Physics
2 answers:
erastova [34]2 years ago
8 0

Answer:

4m/s2

Explanation:

The following data were obtained from the question:

U (initial velocity) = 10m/s

V (final velocity) = 30m/s

t (time) = 5secs

a (acceleration) =?

Acceleration is the rate of change of velocity with time. It is represented mathematically as:

a = (V - U)/t

Now, with this equation i.e

a = (V - U)/t, we can calculate the acceleration of the race car as follow:

a = (V - U)/t

a = (30 - 10)/5

a = 20/5

a = 4m/s2

Therefore, the acceleration of the race car is 4m/s2

VLD [36.1K]2 years ago
8 0

Answer:

a = 4\,\frac{m}{s^{2}}

Explanation:

The acceleration of the racecar is:

a = \frac{v-v_{o}}{\Delta t}

a = \frac{30\,\frac{m}{s} - 10\,\frac{m}{s} }{5\,s}

a = 4\,\frac{m}{s^{2}}

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A football player kicks a football downfield. The height of the football increases until it reaches a maximum height of 15 yards
Trava [24]

Answer:

kick 1 has travelled 15 + 15 = 30 yards before hitting the ground

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards  

Explanation:

1st kick travelled 15 yards to reach maximum height of 8 yards

so, it has travelled 15 + 15 = 30 yards before hitting the ground

2nd kick is given by the equation

y (x) = -0.032x(x - 50)

Y = 1.6 X - 0.032x^2

we know that maximum height occurs is given as

x = -\frac{b}{2a}

y =- \frac{1.6}{2(-0.032)} = 25

and maximum height is

y = 1.6\times 25 - 0.032\times 25^2

y = 20

so kick 2 travels 25 + 25 = 50 yards before hitting the ground

first kick reached 8 yards and 2nd kick reached 20 yards

8 0
2 years ago
Read 2 more answers
A man pushes his child in a grocery cart. The total mass of the cart and child is 30.0 kg. If the force resisting the carts moti
-BARSIC- [3]

The force applied by the man is 60 N

Explanation:

We can solve this problem by applying Newton's second law, which states that:

\sum F = ma (1)

where

\sum F is the net force acting on the child+cart

m is the mass of the child+cart system

a is their acceleration

In this problem, we have:

m = 30.0 kg is the mass

a=1.50 m/s^2

And there are two forces acting on the child+cart system:

  • The forward force of pushing, F
  • The force resisting the cart motion, R = 15.0 N

Therefore we can write the net force as

\sum F = F -R

where R is negative since its direction is opposite to the motion

So eq.(1) can be rewritten as

F-R=ma

And solving for F,

F=ma+R=(30.0)(1.50)+15.0=60 N

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

4 0
2 years ago
If a train is 100 kilometers away, how much sooner would you hear the train coming by listening to the rails (iron) as opposed t
Whitepunk [10]
From tables, the speed of sound at 0°C is approximately
V₁ = 331 m/s (in air)
V₃ = 5130 m/s (in iron)

Distance traveled is
d = 100 km = 10⁵ m

Time required to travel in air is
t₁ = d/V₁ = 10⁵/331 = 302.12 s

Time required to travel in iron is
t₂ = d/V₂ = 10⁵/5130 = 19.49 s

The difference in time is
302.12 - 19.49 = 282.63 s

Answer:  283 s (nearest second)



6 0
2 years ago
Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc
xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0
2 years ago
(d) A beam of white light shines onto a sheet of white paper. An identical beam of light shines onto a mirror. The light is scat
irakobra [83]
QUESTION:-A beam of white light shines onto a sheet of white paper. An identical beam of light shines onto a mirror. The light is scattered from the paper and reflected from the mirror.
Describe how scattering by paper and reflection by a mirror are different from each other.


ANSWER: Scattering sends or reflects light from each point on the object in all directions, whereas reflection sends light from each point on the object in one direction only (or to one point)
5 0
2 years ago
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