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3 years ago

What unbalanced force is needed to give a 976 kg vehicle an acceleration of 2.50 m/s2? ASAP

Physics

1 answer:

RUDIKE [14]3 years ago

4 0

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 976 × 2.5

We have the final answer as

Hope this helps you

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Ok these are my last points say hi if u want brainliest

lawyer [7]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down

soldier1979 [14.2K]

Answer:

$1.9\times 10^{-4}$

$1.2\times 10^{-4}\ m$

Explanation:

r = Radius = 2.7 cm

F = Force = $3.2\times 10^4\ N$

A = Area = $\pi r^2$

$\sigma$ = Stress = $\frac{F}{A}$

E = Young's modulus = $7\times 10^{10}\ Pa$

$\epsilon$ = Strain

$L_0$ = Original length = 67 cm

$\Delta L$ = Change in length

Young's modulus is given by

$E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}$

Strain is $1.9\times 10^{-4}$

Strain is given by

$\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m$

The cylinder height decreases by $1.2\times 10^{-4}\ m$

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3 years ago

An object's movement around an internal axis is _____.

zaharov [31]

Answer:

brooo it is rotation .............

4 0

2 years ago

Any one their to help

Dmitry [639]

Answer:

five dollars

Explanation:

im thick lol

7 0

2 years ago

John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2

Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function $cos(\omega t +\phi)$ is defined between -1 and 1, So it is not possible obtain a value $2\pi$ greater.

In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have $f=1+cos(\omega t +\phi)$ , the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the $f=1+cos(\omega t +\phi)$ , and its mean that the function $f=cos(\omega t +\phi)$ , in general, is not greater than $cos(\omega t +\phi)$ .

3 0

3 years ago