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3 years ago

What is the mass of a shopping cart that is pushed with a net force of 10 N to accelerate at the rate of 0.5 m/s2?

2 answers:

6 0

Actualy the correct answer is going to be 20kg. And this how i got it. All i did was 10 divided by .5 and i got the answer.

Please mark brainiest please.

nekit [7.7K]3 years ago

3 0

20kg give homie his brainliest though he earned it

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You are given two converging lenses to build a compound microscope. Lens A has focal length 0.50 cm and lens B has focal length

siniylev [52]

Answer:

The lens to be used for the objective is lens A

Explanation:

The objective of a compound microscope

The focal length of the lens used for the objective = 1/(magnification obtained)

The focal length of most modern is equal to the tube length

The range of sizes for the focal length of a microscope is between 2 mm and 40 mm

Therefore, the appropriate lens to be used for the objective of the compound is lens A that has a focal length of 0.50 cm = 5 mm

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3 years ago

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ipn [44]

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3 years ago

What is the term used to label the energy levels of electrons?

mel-nik [20]

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3 years ago

Calculate the object's velocity as shown on the position-time graph,

Cloud [144]

Answer:

10 m/s

Explanation:

The following data were obtained from the question:

Initial Displacement (d1) = 10 m

Final Displacement (d2) = 60 m

Initial time (t1) = 0 s

Final time (t2) = 5 s

Velocity (v) =.?

Next, we shall determine the change in displacement of the object and likewise the change in time.

This can be obtained as follow:

Initial Displacement (d1) = 10 m

Final Displacement (d2) = 60 m

Change in displacement (Δd) = d2 – d1

Change in displacement (Δd) = 60 – 10

Change in displacement (Δd) = 50 m

Initial time (t1) = 0 s

Final time (t2) = 5 s

Change in time (Δt) = t2 – t1

Change in time (Δt) = 5 – 0

Change in time (Δt) = 5 s

Finally, we shall shall calculate the velocity of the object as illustrated below:

Change in displacement (Δd) = 50 m

Change in time (Δt) = 5 s

Velocity (v) =.?

v = Δd/Δt

v = 50/5

v = 10 m/s

Therefore, the velocity of the object is 10 m/s.

4 0

3 years ago

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/mN/m. At t=0 t=0 the block has velocity -

slega [8]

Answer:

The amplitude of the spring is 32.6 cm.

Explanation:

It is given that,

Mass of the block, m = 2 kg

Force constant of the spring, k = 300 N/m

At t = 0, the velocity of the block, v = -4 m/s

Displacement of the block, x = 0.2 mm = 0.0002 m

We need to find the amplitude of the spring. We know that the velocity in terms of amplitude and the angular velocity is given by :

$v=\omega\sqrt{A^2-x^2}$

$\omega=\sqrt{\dfrac{k}{m}}$

$\omega=\sqrt{\dfrac{300}{2}}$

$\omega=12.24\ rad/s$

So, $\dfrac{v^2}{\omega^2}+x^2=A^2$

$\dfrac{(-4)^2}{(12.24)^2}+(0.0002)^2=A^2$

A = 0.326 m

A = 32.6 cm

So, the amplitude of the spring is 32.6 cm. Hence, this is the required solution.

4 0

3 years ago