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2 years ago

A rollercoaster accelerates from 10 m/s to 100 m/s2 for 25 seconds. What is the acceleration?

Physics

1 answer:

UkoKoshka [18]2 years ago

5 0

Answer:

A roller coasters accelerates from an initial velocity of of 6.0 m/s to a final velocity of 70 m/s over 4 seconds. What's the acceleration? Q. Acceleration only takes place when things speed up. Q. A drag racer accelerated from 0 m/s to 200 m/s in 5 s.

Explanation:

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What are the 4 energy transformations of a dash toy

Hunter-Best [27]

Answer:

Thermal/Heat energy, kinetic energy, light energy, & Electromagnetic energy

3 0

2 years ago

A 7-kg bowling ball moving at 4 m/s strikes a 1 kg bowling pin. If the ball slows to 2 m/s in 0.05 s, how much force does it exe

iren [92.7K]

Answer:

280 N

Explanation:

acceleration = v2-v1 / time taken = (2-4 )/ 0.05 = -40 m/s^2 ( neg sign indicates slowing down )

force exerted = ma = 7 kg x -40 m/s^2 = - 280 N ( neg sign means opposite direction of initial velocity )

since the 7 kg ball is slowing down, the direction of force will be opposite of the initial velocity , and it will be 280 N

7 0

2 years ago

Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

velocity of wave on the string with greater tension;

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

Fundamental frequency of wave on the string with greater tension;

$f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$

Beat frequency = F₂ - F₁

= 270.6 - 258

= 12.6 Hz

Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0

3 years ago

The part of the electromagnetic spectrum you can see is called ultraviolet light. Please select the best answer from the choices

Artyom0805 [142]

True because when the light bounces off a surface it creates a rainbow which we call roygbiv

4 0

3 years ago

A motorcycle is travelling north at 11m/s. How much time would it take for the motorcycle to increase its velocity to 26m/s (N)

DaniilM [7]

Using the kinematic equation,
v = u + at
26 = 11 + 3t
15 = 3t
t = 5 seconds.
Time required is 5 seconds.

8 0

3 years ago