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Drupady [299]
3 years ago
6

A car is going 3 m/s, then it speeds up for 4 seconds. If it has an acceleration of 5m/s^2, find its final velocity. Which equat

ion can you use?
A Δx = vi*t + (½)at^2
B vf = vi + a*t
C Δx = (½)(vf + vi)*t
D vf^2 = vi^2 + 2aΔX
Physics
1 answer:
andreyandreev [35.5K]3 years ago
4 0

Well, since the question GIVES you the initial velocity, the acceleration, and the time, and ASKS for the final velocity, you'd be smart to find an equation that USES the initial velocity, the acceleration, and the time, and FINDS the final velocity.

Have a look at equation B ..... vf = vi + a*t . That's pretty durn close !

vf = (initial velocity) + (acceleration)*(time)

vf = (3 m/s) + (5 m/s²)*(4 sec)

vf = (3 m/s) + (20 m/s)

vf = 23 m/s

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3 years ago
you slide abox of books at constant speed up a30 degree ramp, applying a force of 200 newtons directed up the slope. The coeffic
Tomtit [17]

The work done is 400 J

Explanation:

The work done by you in pushing the box along the slope is given by

W=Fd

where

F is the magnitude of the force applied

d is the distance covered by the box along the slope

Here we have the following:

F = 200 N is the magnitude of the force applied

d=\frac{h}{sin \theta} is the distance covered, where

h = 1 m is the vertical rise

\theta=30^{\circ} is the slope of the plane

Substituting and solving, we find

W=F\frac{h}{sin \theta}=\frac{(200)(1)}{sin 30^{\circ}}=400 J

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5 0
3 years ago
Suppose electric power is supplied from two independent sources which work with probabilities 0.4, 0.5, respectively. if both so
sattari [20]

Answer:

The answers to the questions are as follows

a)  k = 0, P = 0.3

k = 1, P = 0.5

k = 0, P = 0.2

b) The probability that enough power will be available is 0.5.

Explanation:

To solve the question we write the parameters as follows

Probability that the first power source works = P(A) = 0.4

Probability that the second power source works = P(B) = 0.5

When both sources are supplying power  we have the probability = 1

If non of them is producing the probability = 0

a) The probability that exactly k sources work for k=0,1,2 is given by

For k = 0, probability = (1- P(A))× (1- P(B)) = 0.6 × 0.5 =0.3

Therefore the probabilities that exactly 0 source work  = 0.3

for k = 1 we have the probability = P(A)(1-P(B)) + P(B)(1-P(A)

= 0.4(1-0.5)+0.5(1-0.4) = 0.2 + 0.3 = 0.5

The probabilities that exactly 1 source work  = 0.5

for k = 2 we have the probability given by = P(A) × P(B) = 0.4 × 0.5 = 0.2

Therefore the  probability that exactly 2 sources work  = 0.2

b)  The probability that enough power will be available is

0 × P(k = 0) + 0.6 × P(k = 1) + 1 × P(k = 2)

0 × 0.2 + 0.6 × 0.5 + 1 × 0.2 = 0.5

The probability that enough power will be available is 0.5.

3 0
3 years ago
What happens in a tug of war if the net forces are balanced and why?
FinnZ [79.3K]

Answer:

Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.

Explanation:

6 0
3 years ago
A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple
aliina [53]

Answer:

T= 1 s

Explanation:

Given that

When x=  cm ,T= 1

we know that time period of spring mas system given as

T=2\pi \sqrt{\dfrac{m}{k}}

T= Time period

m= mass

k=spring constant

So from above equation we can say that time period of system does not depends on the value of x.

So when x= 10 cm ,still time period will be 1 s.

T= 1 s

5 0
3 years ago
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