Ask question

Ask question

All categories

4 years ago

7

If weight is a measure of an object's force due to gravity, what is the weight of a student who has a mass of 65 kg? (assume tha

t the acceleration due to gravity equals 9.8 m/s2.)

1 answer:

Anton [14]4 years ago

6 0

The weight will be around 637N

F=mg

You might be interested in

A black widow spider hangs motionless from a web that extends vertically from the ceiling above. If the spider has a mass of 1.2

leva [86]

Answer:

Tension = 0.012 N

Explanation:

If the black widow spider is hanging vertically motionless from the ceiling above. Then, the weight of the spider must be balancing the tension in the spider web. Therefore,

Tension = Weight

Tension = mg

where,

m = mass of spider = 1.27 g = 0.00127 kg

g = acceleration due to gravity = 9.8 m/s²

Therefore,

Tension = (0.00127 kg)(9.8 m/s²)

<u>Tension = 0.012 N</u>

8 0

3 years ago

Which one of the following substances is a liquid fuel used in rocket engines?

melomori [17]

There are none on the list you included with your question.

8 0

4 years ago

Read 2 more answers

A box of volume V has a movable partition separating it into two compartments. The left compartment contains 3000 particles, the

storchak [24]

Answer:

a) V1 = 4V - V2/3 and V2 = 4V - 3V1

b) Δe = 4000V - 4000V2 + 9000V1

Explanation:

Let V represent volume of the box containing the two compartments

V1 represents compartment of the left compartment

V2 represents compartment of the right compartment

Momentum of the compartments before impact:

3000V1 + 1000V2

Momentum of the compartments after impact:

V(3000 + 1000) = 4000V

a) To obtain the volume of each compartment, that is, V1 and V2, we say:

Momentum before impact = Momentum after impact

3000V1 + 1000V2 = 4000V

∴ V1 = 4000V - 1000V2/3000 = 4V - V2/3

Also, V2 = 4000V - 3000V1/1000 = 4V - 3V1

b) Change in entropy,Δe = 4000V1 - 1000V2

By substituting the V1 and V2, we have:

4000(4V - V2)/3 - 1000(4V - 3V1)

16000V - 4000V2/3 - 4000V + 3000V1

16000V - 4000V2 - 12000V + 9000V1

∴ Δe = 4000V - 4000V2 + 9000V1

6 0

3 years ago

The force required to stretch a Hooke’s-law spring varies from 0 N to 63.5 N as we stretch the spring by moving one end 5.31 cm

Alika [10]

Answer:

Force constant will be 1195.85 N/m

Work done will be 1.6859 J

Explanation:

We have given the force, F = 63.5 N

Spring is stretched by 5.31 cm

So x = 0.0531 m

Force is given , F = 63.5 N

We know that force is given by $F=kx$

So $63.5=k\times 0.0531$

k = 1195.85 N/m

Now we have to find the work done

We know that work done is given by

$W=\frac{1}{2}kx^2=\frac{1}{2}\times 1195.85\times 0.0531^2=1.6859J$

8 0

3 years ago

A boy on a skateboard is pushed down the sidewalk by a friend. The skateboard rolls for a while but eventually comes to a stop.

Ket [755]

Friction is causing the skateboard to stop rolling.

4 0

3 years ago

Read 2 more answers