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3 years ago

Claudia stubs her toe on the coffee table with a force of 100.ON. A) What is the

Physics

1 answer:

slamgirl [31]3 years ago

7 0

Answer:

A.) acceleration= 55.6m/s^2

B.) acceleration of table= 5.0m/s^2

C.) More acceleration

Explanation:

A.) 100N/1.8kg= -55.6

B.) 100N÷20kg= 5

C.) Because since the table would have less mass, it would have had to accelerate more

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4 years ago

the human nervous system can propagate nerve impulses at about 10 squared metered per second. estimate the time it takes to trav

rusak2 [61]

Use the formular d = v x t
d = 2m
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4 years ago

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3 years ago

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A block is on a frictionless table, on earth. The block accelerates at 7.5 m/s when a 70 N horizontal force is applied to it. Th

Liono4ka [1.6K]

Answer:

The weight of the block on the moon is 15 kg.

Explanation:

It is given that,

The acceleration of the block, a = 7.5 m/s²

Force applied to the box, F = 70 N

The mass of the block will be, $m=\dfrac{F}{a}$

$m=\dfrac{70\ N}{7.5\ m/s^2}$

m = 9.34 kg

The block and table are set up on the moon. The acceleration due to gravity at the surface of the moon is 1.62 m/s². The mass of the object remains the same. It weight W is given by :

$W=m\times g$

$W=9.34\ kg\times 1.62\ m/s^2$

W = 15.13 N

W = 15 N

So, the weight of the block on the moon is 15 kg. Hence, this is the required solution.

3 0

3 years ago

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago