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3 years ago

How do you calculate acceleration using a friction coefficient?

1 answer:

4 0

Use the diagram to determine the normal force, the net force, the coefficient of friction (μ) between the object and the surface, the mass, and the acceleration of the object. (Neglect air resistance.) Note: To simplify calculations, an approximated value of g is often used - 10 m/s/s.

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A force of 6.00 N acts in the positive direction on a 3.00 kg object, originally traveling at +15.0 m/s, for 10.0 s. (a) What is

frozen [14]

Answer:

60 kg m/s

Explanation:

Let $a\;\; m/s^2$ be the acceleration of the object.

As the acceleration of the object is constant, so

$a=\frac {v-u}{t}\cdots(i)$

Given that applied force, F=6.00 N,

From Newton's second law, we have

$F= m\times a$ ,

$\Rightarrow F=\frac {m(v-u)}{t}$ [from equation (i)]

$\Rightarrow Ft=m(v-u)$

$\Rightarrow Ft=mv-mu$

$\Rightarrow mv-mu=6\times 10$ [given that time, t=10 s and F=6 N]

$\Rightarrow mv-mu=60 kg \;m/s$

Here mv is the final momentum of the object and mu is the initial momentum of the object.

So, the change in the momentum of the object is mv-mu.

Hence, the change in the momentum of the object is 60 kg m/s.

6 0

3 years ago

Encuentre en km.h la velocidad de un tigre que corre 550 km.h en 69min

Len [333]

Answer:

v = 478.26 km/h

Explanation:

The question is "find in km.h the speed of a tiger that runs 550 km in 69min"

Distance, d = 550 km

Time, t = 69 min = 1.15 h

We need to find the speed of the tiger. The speed of an object is equal to the total distance covered divided by time taken. So,

$v=\dfrac{550\ km}{1.15\ h}\\\\v=478.26\ km/h$

So, the speed of the tiger is 478.26 km/h.

8 0

3 years ago

Is Time and Light both connected in anyway?

MissTica

Answer:

It is not connected by any way

8 0

2 years ago

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles

Anna007 [38]

Question

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles travelled in a straight line and some were deflected at different angles.

Which statement best describes what Rutherford concluded from the motion of the particles?

A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.

B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.

C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.

D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Answer:

The right answer is C)

Explanation:

In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.

So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".

Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.

Cheers!

6 0

2 years ago

Read 2 more answers

What does the change of motion of an object depend on?

IrinaVladis [17]

Answer:

I think the answer is B

Explanation:

4 0

3 years ago

Read 2 more answers