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jeyben [28]
2 years ago
11

18. How much force would it take to push another, larger friend who has a mass of 70 Kg to accelerate

Physics
1 answer:
oksano4ka [1.4K]2 years ago
4 0

Answer:

280N

Explanation:

f = mass x acceleration

You might be interested in
A 5.0 kg object moving at 10 m/s on a frictionless surfaces collides with but does not stick to a 2.0 kg object that is initiall
lilavasa [31]

Answer:

I= 20 i {N.s}

Explanation:

In order to obtain the impulse on the 2 kg ball, you have to apply the equation of Impulse:

I=FΔt

Where I is the impulse vector, F is the net force and Δt is the interval of time when the force is applied.

In this case:

Δt=0.01 s

F= 2000 i N

where i is the unit vector in the x direction.

Replacing the values in the formula:

I=(2000)(0.01)i

Therefore:

I= 20 i {N.s}

3 0
3 years ago
A water bottle is dropped from a plane and accelerates because of gravity. What distance did it fall during 3 seconds?
exis [7]
It’d fall 29.4m or 96.46ft

Explanation:
Uhh since gravity is 9.8m/s then in three seconds it’d drop 29.4m or 96.46ft
That is assuming there isn’t a lot of wind resistance, but if you take that into account, then it’d probably be somewhere around 25m since the water bottle is going to be heavier than the wind resistance, and since we don’t know the weight of the water bottle it can’t really be calculated.

Hope this helps!
4 0
2 years ago
1. Viruses are unique among infectious agents because they are
yuradex [85]
1,) C

2,) C

Hope this helps
4 0
3 years ago
A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

\beta=8.86\times10^{9}

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

Put the value into the formula

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

T=7.74\times10^{-8}

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

6 0
3 years ago
Hii please help i’ll give brainliest if you give a correct answer please please hurry
fiasKO [112]

Answer:

B most likely

Explanation:

tell me if this is right or wrong

8 0
2 years ago
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