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3 years ago

Janet pushes a cart, the cart with a net force of 60N. The mass of the cart was 48 kg. Calculate the carts acceleration

Physics

1 answer:

jeka57 [31]3 years ago

5 0

Answer:

a = 1.25m/s^2

Explanation:

Force = mass x acceleration

Given mass M = 48kg and

Force F = 60N

60 = 48 x a

Divide both sides by 48

60/48 = 48 x a/48

1.25 = a

a = 1.25m/s^2

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If object A has more mass than object B, what will object A need to accelerate at the same rate as object B?

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Answer:

More force

Explanation:

Object A has more mass than object B

For object A to accelerate at the same rate as object B, it will need more force.

According to Newton's second law of motion "the net force on a body is the product of its mass and acceleration".

Net force = mass x acceleration

Now, if a body has more mass and needs to accelerate at the same rate as another one with a lower mass, the force on it must be increased.

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A ball is dropped from rest at the top of a 6.10 m

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Answer:

n = 5 approx

Explanation:

If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back

$\frac{v_1}{v}$ = e ( coefficient of restitution ) = $\frac{1}{\sqrt{10} }$

and

$\frac{v_1}{v} = \sqrt{\frac{h_1}{6.1} }$

h₁ is height up-to which the ball bounces back after first bounce.

From the two equations we can write that

$e = \sqrt{\frac{h_1}{6.1} }$

$e = \sqrt{\frac{h_2}{h_1} }$

So on

$e^n = \sqrt{\frac{h_1}{6.1} }\times \sqrt{\frac{h_2}{h_1} }\times... \sqrt{\frac{h_n}{h_{n-1} }$

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Taking log on both sides

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3 years ago

Which is a component of pseudoscience, but not science?

quester [9]

Answer:

The statements, observations, beliefs and suppositions all are the components of the pseudoscience.

Explanation:

The claims included in the pseudoscience including the beliefs, statements and practices are claimed to be scientific but these are devoid of any scientific evidences provided by the experimental procedures.

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3 years ago

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dlinn [17]

Explanation:

Given that,

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Using formula of energy

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$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-9}}$

$E=3.315\times10^{-17}\ J$

(b). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-9})^2}$

$E=6.709\times10^{-21}\ J$

(c). We need to calculate the energy of photon

Using formula of energy

$E = \dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{6.0\times10^{-15}}$

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(d). We need to calculate the kinetic energy of an electron

Using formula of kinetic energy

$\lambda=\dfrac{h}{\sqrt{2mE}}$

$E=\dfrac{h^2}{2m\lambda^2}$

Put the value into the formula

$E=\dfrac{(6.63\times10^{-34})^2}{2\times9.1\times10^{-31}\times(6.0\times10^{-15})^2}$

$E=6.709\times10^{-9}\ J$

Hence, This is the required solution.

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Where are the answer choice ?

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