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MakcuM [25]
4 years ago
15

What is the value of acceleration that the car experiences

Physics
1 answer:
Anni [7]4 years ago
7 0
Can u show the problem ?
Acceleration= the rate and change of an object also cars go fast so I think it would be positive acceleration
Hope I helped
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A bridge that was 5.0 m long has been washed out by the rain several days ago. How fast must a car be going to successfully jump
amid [387]

Answer:

Speed, v = 7.83 m/s

Explanation:

It is given that,

Length of the bridge, l = 5 m

The road on the far side is 2.0 m lower than the road on this side, x = 2 m

The horizontal distance covered by the car is 5 meters and the vertical distance covered by the car is 2 meters.

Initial speed of the car, u = 0

Let t is the time taken by the car . Using the second equation of motion as :

For vertical distance :

s=ut+\dfrac{1}{2}gt^2

t=\sqrt{\dfrac{2\times 2}{9.8}}=0.638\ s

Let v is the velocity to jump the stream. It is given by :

Horizontal distance, d = 5 m

v=\dfrac{5\ m}{0.638\ s}

v = 7.83 m/s

So, the car should travel with a speed of 7.83 m/s. Hence, this is the required solution.

8 0
4 years ago
The distance versus time plot for a particular object shows a quadratic relationship. Which column of distance data is possible
Vladimir79 [104]
The correct choice is (C)

8 0
3 years ago
Read 2 more answers
9. A van travels at a speed of 40.0 m/s has a mass of 600kg. What is the van’s kinetic energy? *
Zina [86]

ke =  \frac{1}{2} m{v}^{2}  \\  =  \frac{1}{2} (600)( {40}^{2} ) \\  = 300 \times 800 = 240000 = 240kjoule

8 0
3 years ago
A mass m at the end of a spring vibrates with a frequency
Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

F = -kx

And Newton's Second Law also states that

F = ma = m\frac{d^2x}{dt^2}

Combining two equations yields

a = -\frac{k}{m}x

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

\omega = \sqrt{\frac{k}{m}}

And given that ω = 2πf

the relation between frequency and mass becomes

f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}.

Let's apply this to the variables in the question.

0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg

6 0
3 years ago
Compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite having a 0.25 volume fraction of fibers, ass
zmey [24]

Answer:

632.5 MPa

Explanation:

\sigma_{m} = Matrix stress at fiber failure = 10 MPa

V_f = Volume fraction of fiber = 0.25

\sigma_f = Fiber fracture strength = 2.5 GPa

The longitudinal strength of a composite is given by

\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa

The longitudinal strength of the aligned carbon fiber-epoxy matrix composite is 632.5 MPa

8 0
3 years ago
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