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stellarik [79]
2 years ago
15

If a Ferrari, with an initial velocity of 10 m/s, accelerates at a rate of 50 m/s 2 for 3 seconds, what will its final velocity

be? Use the following formula:
Variable
Equation
Solve

Physics
2 answers:
Evgen [1.6K]2 years ago
3 0

Answer:

160m/s

Explanation:

The Ferrari is moving by uniformly accelerated motion, with constant acceleration of a = 50 m/s^2, and initial velocity u = 10 m/s. The velocity at time t of the car is given by

v(t)=u+at

where

u = 10 m/s

a = 50 m/s^2

If we substitute t = 3 s into the equation, we can find the velocity of the car after 3 seconds:

v(3s) = 10m/s + (50m/s^2)(3s)=160m/s

ValentinkaMS [17]2 years ago
3 0

Answer:

160m/s

Explanation:

<em>Rearrange</em><em> the formula:</em>

acceleration = vf - vi ÷ time

acceleration × time = vf - vi

(acceleration × time) + vi = vf

<em>The values are:</em>

acceleration = 50m/s²

time = 3s

vi = 10 m/s

<em>Substitute the values into the formula</em>

(50 × 3) + 10 = 160m/s

The final velocity = 160m/s

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problems like this A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet fi
Ymorist [56]

This question is incomplete the complete question is

A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?

Answer:

(a) Xs=0.459m

(b) t=0.984 s

(c) Vc=6.65 m/s

Explanation:

(a) To reach maximum distance

g=-9.8m/s^{2}\\ Vf=0\\v_{b}^{2}=v_{a}^{2}+2gx_{s} \\  x_{s}=\frac{0-(3^{2} )}{-2*9.8}\\ x_{s}=0.459m

(b) For Time

To find t we must find t1 and t2

as

t=t1+t2

For T1

t_{1}=(Vb-Va)/g \\t_{1}=(0-3)/9.8\\t_{1}=0.306s

For T2

x_{l}=Vbt+(1/2)gt_{2}^{2}\\   as\\x_{l}=x_{1}+x_{s}\\x_{l}=1.8+0.459\\x_{l}=2.259\\so\\t_{2}=\frac{2.259*2}{9.8} \\t_{2}=0.6789s

For Total Time

t=t1+t2

t=0.306+0.6789

t=0.984s

(c) To find Vc

Vc=Vb+gt2

Vc=(0)+(9.8)(0.6789)

Vc=6.65 m/s

7 0
2 years ago
1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
2 years ago
Why does damp soil help excess electrons move
Marina CMI [18]

Explanation:

  • Because where there is water, energy is conducted, and <u>moves </u>the electrons.
  • They also move excess electrons because pure water is a poor conductor

3 0
3 years ago
The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p
NISA [10]

Answer:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be v_{s} = 1.5 m/s

The separation of the two sides of the river, d = 80 m

The time taken by the swimmer to get to the other end of the river bank,

t = \frac{d}{v_{s} }

t = 80/1.5

t = 53.33 s

The swimmer will be carried downstream by the river through a distance, s

Let the velocity of the river be v_{r} = 2.5 m/s

S = v_{r} t

S = 53.33 * 2.5

S = 133.33 m

c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

That is ,

cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

d) Downstream velocity of the swimmer, v_{y} = v_{s} sin \theta\\

v_{y} = 1.5 sin 53.13\\v_{y} = 1.2 m/s

The vertical displacement is given by, y = v_{y} t

80 = 1.2 t

t = 80/1.2

t = 66.67 s

the horizontal speed,

v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s

The downstream horizontal distance of the swimmer, x = v_{x} t

x = 1.6 * 66.67

x = 106.67 m

7 0
3 years ago
the rock of 10 kg is falling near the Earth's surface.assume that g =10N/kg and the is no air resistance. what is the accelerati
Ivanshal [37]

Answer:

kya faltu sawal h repetitive g=10N/kg

5 0
2 years ago
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