Ask question

Ask question

All categories

3 years ago

7

A 52.3 kg student is standing at rest on roller skates when another student throws a bag of 3.5 kg bag of sand. After the studen

t catches the bag they are pushed a distance of 2 meters in 1 second. What was the initial velocity of the sand bag being thrown?

1 answer:

IRINA_888 [86]3 years ago

4 0

PLEASE HELP, I need help too !! Please !

Explanation:

You might be interested in

An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. Howeve

Vlad1618 [11]

Answer:

$v=0.9833\ c$

Explanation:

The density changes means that the length in the direction of the motion is changed.

Therefore,

$\text{Density} = \frac{m}{lwh}$

Given :

Side, b = h = 0.13 m

Mass, m = 3.3 kg

Density = 8100 $kg/m^3$

So,

$8100=\frac{3.3}{l \times 0.13 \times 0.13}$

$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$

l = 0.024 m

Then for relativistic length contraction,

$l= l' \sqrt{1-\frac{v^2}{c^2}}$

$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$

$0.184= \sqrt{1-\frac{v^2}{c^2}}$

$0.033= 1-\frac{v^2}{c^2}}$

$\frac{v^2}{c^2}= 0.967$

$\frac{v}{c}=0.9833$

$v=0.9833\ c$

Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).

3 0

3 years ago

Calculate the workdone to stretch an elastic string by 40cm if a force of 10 newton produces an extension of 4cm in it

notka56 [123]

Answer:

Workdone = 20 Joules

Explanation:

Given the following data;

Force = 10N

Extension, e = 4cm to meters = 4/100 = 0.04 meters

Workdone extension = 40cm to meters = 40/100 = 0.4 meters

To find the work done;

First of all, we would find the spring constant using the formula;

Force = spring constant * extension

10 = spring constant * 0.04

Spring constant = 10/0.04

Spring constant = 250 N/m

Next, we find the work done;

Workdone = ½ke²

Where;

k is the spring constant.

e is the extension.

Substituting into the formula, we have;

Workdone = ½ * 250 * 0.4²

Workdone = 125 * 0.16

Workdone = 20 Joules

5 0

3 years ago

A new mechanic foolishly connects an ammeter with 0.1 Ω resistance directly across a 12-V car battery with internal resistance o

NemiM [27]

Answer:

The power dissipated by the meter is 1188W

Explanation:

Here we have a circuit constituted with a power source and two resistors in series, we can calculate the power dissipated by the meter using the following formula:

$P=I^2R_m$

We first need to fin the current going through the circuit:

$I=\frac{V}{R}\\where:\\V=voltage\\R=resistance$

$R=R_s+R_m$

because they are connected in series. So:

$I=\frac{12V}{(0.01+0.1)}=109A$

$P=(109)^2*0.1=1188W$

8 0

3 years ago

Two charges are sitting 1.5 m apart with a force of 3 N between them. They are now moved farther apart to 2.25 m and one of the

ANTONII [103]

Answer: 2.37N

Explanation:

According to coulombs law which states that the force of attraction (F) between two charges (q1 and q2) is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them. Mathematically,

F = kq1q2/r²

For the first two charges that are sitting 1.5 m apart with a force of 3 N between them, we have

3 = kq1q2/1.5²

3 = kq1q2/2.25

Kq1q2= 6.75... (1)

If the charges are now moved farther apart 2.25 m and one of the charges is increased by a factor of 4. The formula becomes

F2 = k(4q1)q2/2.25² (q1 has been increased by factor of 4)

k(4q1)q2 = 5.06F2 ... (2)

Dividing 2 by 1 we have

k(4q1)q2/kq1q2 = 5.06F2/3

4 = 5.06F2/3

5.06F2 = 12

F2= 12/5.06

F2 = 2.37N

Therefore the magnitude of the new force between the two charges is 2.37N

5 0

3 years ago

What is the net displacement of the particle between 0 seconds and 80seconds?

Ira Lisetskai [31]

Answer:

Option (A)

Explanation:

Displacement of a particle on a velocity time graph is represented by the area between the line representing velocity and x-axis (time).

Displacement of a particle from t = 0 o t = 40 seconds = Area of ΔAOB

Area of triangle AOB = $\frac{1}{2}(\text{Base})(\text{height})$

= $\frac{1}{2}(40)(4)$

= 80 m

Similarly, displacement of the particle from t = 40 to t = 80 seconds = Area of ΔBCD

Area of ΔBCD = $\frac{1}{2}(40)(4)$

= 80 m

Total displacement of the particle from t = 0 to t = 80 seconds,

= 80 + 80

= 160 m

Option (A) will be the answer.

7 0

3 years ago