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Alexandra [31]
3 years ago
8

A car is traveling at 7.1m/s accelerates 4.3 m/s^2 to reach a speed of 16.0 m/s how long does it take for this acceleration to o

ccur
Physics
1 answer:
Bond [772]3 years ago
4 0

<u>Answer</u>

2.06977 seconds


<u>Explanation</u>

Acceleration, a, is the rate of change of velocity

a = (v-u)/t

  where v and u are final and initial velocities respectively.

4.3 = (16.0 - 7.1)/t

4.3 = 8.9/t

t = 8.9/4.3

   = 2.06977 seconds

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Stels [109]

Answer:

<h2><em><u>atoms can form new substances </u></em></h2>
5 0
3 years ago
a 3 kg piece of putty that is moving with a velocity of 10 m/s collides and sticks to an 8 kg bowling ball that was at rest. wha
defon

The final velocity is 2.7 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.

Therefore we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 = 3 kg is the mass of the putty

u_1 = 10 m/s is the initial velocity of the putty (we take its direction as positive direction)

m_2 = 8 kg is the mass of the ball

u_2 = 0 m/s is the initial velocity of the ball (at rest)

v is the final combined velocity of the two putty+ball

Re-arranging the equation and substituting the values, we find the  final combined velocity:

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(3)(10)+0}{3+8}=2.7 m/s

And the positive sign indicates their final direction is the same as the initial direction of the putty.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

3 0
3 years ago
GIVING BRAINLIEST PLEASE HELP!!
levacccp [35]

Answer: wood

Explanation: pls help me brainly.com/question/22777864

7 0
3 years ago
A 200 g block is pressed against a spring of force constant 1.40 kN/m until the block compresses the spring 10.0 cm. The spring
tino4ka555 [31]

Answer:

Explanation:

This problem bothers on the energy stored in a spring in relation to conservation of energy

Given data

Mass of block m =200g

To kg= 200/1000= 0.2kg

Spring constant k = 1.4kN/m

=1400N/m

Compression x= 10cm

In meter x=10/100 = 0.1m

Using energy considerations or energy conservation principles

The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring

Potential Energy stored in spring

P.E=1/2kx^2

Kinetic energy of the block

K.E =1/mv^2

Where v = velocity of the block

K.E=P.E (energy consideration)

1/2kx^2=1/mv^2

Kx^2= mv^2

Solving for v we have

v^2= (kx^2)/m

v^2= (1400*0.1^2)/0.2

v^2= (14)/0.2

v^2= 70

v= √70

v= 8.36m/s

a. Distance moved if the ramp exerts no force on the block

Is

S= v^2/2gsinθ

Assuming g= 9. 81m/s^2

S= (8.36)^2/2*9.81*sin60

S= 69.88/19.62*0.866

S= 69.88/16.99

S= 4.11m

6 0
3 years ago
What is the displacement of the runner, whose velocity versus time graph is shown in the Figure, in the first 15.5 s?
muminat

Answer:

10 displacement of the runner

6 0
3 years ago
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