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4 years ago

How do you calculate mass with distance of 30m and kinetic energy 4392j

1 answer:

4 0

Use the formula, $E_{k} =Fd$ to answer. We need the Mass, which is weight divided by acceleration. So, let's solve for F (force):
$F= \frac{ E_{k} }{d}$

Plug in known values:
$F= \frac{4392}{30}$

Solve:
$F=146.4N$

Then, since we need mass, assuming we are on Earth, and there are no neutron stars, or anything like that close by, we can divide the weight by acceleration of gravity (9.8 m/s^2) to get the mass:
$m= \frac{146.4}{9.8}$

Simplify:
$m=14.94kg$

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Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow

Inessa [10]

The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.

The force per unit length between two parallel thin current-carrying $I_1$ and $I_2$ wires at distance ' r ' is given by $f=\frac{u_0I_1I_2}{2\pi r}$ ....(1) .

If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

$F_2_1=F_2_3$

Using equation (1) , we get

$\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A$

I₃ = 2.4 A and the current is pointing in the downward direction

Learn more about the magnitude and direction of forces here:

brainly.com/question/14879801?referrer=searchResults

#SPJ4

5 0

2 years ago

A 48.0-kg astronaut is in space, far from any objects that would exert a significant gravitational force on him. He would like t

marusya05 [52]

Answer:

The astronaut is moving at a speed of 0.36m/s

Explanation:

Speed here corresponds to velocity

The astronaut's mass = 48kg

velocity of astronaut = ?

mass of socket = 0.72kg

velocity of socket = 5m/s

mass of the spanner = 0.8kg

velocity of spanner = 8m/s

change in time = 0.05 -0 = 0.05sec

mass of the mallet = 1.2kg

velocity of mallet = 6m/s

change in time = 9.9 -0 = 9.9sec

To find the astronaut velocity, we would calculate the total momentum which is the astronaut.

∑momentum (M) = ∑astronaut momentum

∑M = ∑astronaut M

∑astronaut M = M of socket + M of spanner + M of mallet

momentum = mass × velocity

(mass × velocity)of astronaut = (0.72×5) + (0.8×8) + (1.2×6)

48 × velocity of astronaut= 3.6 + 6.4 + 7.2

48 × velocity of astronaut= 17.2

velocity of astronaut = 17.2/48

velocity of astronaut = 0.36m/s

The astronaut is moving at a speed of 0.36m/s

5 0

3 years ago

A child is swinging back and forth with a constant period and amplitude. Somewhere in front of the child, a stationary horn is e

Amanda [17]

Answer:

Explanation:

We shall apply concept of Doppler's effect of apparent frequency to this problem . Here observer is moving sometimes towards and sometimes away from the source . When observer moves towards the source , apparent frequency is more than real frequency and when the observer moves away from the source , apparent frequency is less than real frequency . The apparent frequency depends upon velocity of observer . The formula for apparent frequency when observer is going away is as follows .

f = f₀ ( V - v₀ ) / V , f is apparent , f₀ is real frequency , V is velocity of sound and v is velocity of observer .

f will be lowest when v₀ is highest .

velocity of observer is highest when he is at the equilibrium position or at middle point .

So apparent frequency is lowest when observer is at the middle point and going away from the source while swinging to and from before the source of sound .

3 0

3 years ago

A rope of negligible mass passes over a uniform cylindrical pulley of 1.50kg massand 0.090m radius. The bearings of the pulley h

boyakko [2]

Answer:

Explanation:

mass of pulley, m3 = 1.5 kg

Radius of pulley, R = 0.09 m

mass of monkey, m2 = 4.5 kg

mass of banana bunch, m1 = 3 kg

Let a is teh acceleration ans T1 and T2 be the tension in the rope.

The moment of inertia of the pulley

I = 0.5 x m3 x R² = 0.5 x 1.5 x 0.09 x 0.09 = 0.006075 kgm²

According to Newton's second law

T1 - m1 g = m1 x a .... (1)

m2 g - T2 = m2 x a ..... (2)

(T2 - T1 ) x R = I x α

where, α is the angular acceleration

α = a / R

(T2 - T1)R = 0.5 x m3 x R² x a / R

T2 - T1 = 0.5 x m3 x a ..... (3)

from (1), (2) and (3)

$a = \frac{m_{2}-m_{1}}{m_{1}+m_{2}+\frac{m_{3}}{2}}\times g$

$a = \frac{4.5-3}{3+4.5+0.75}\times 9.8$

a = 1.78 m/s²

from equation (1)

T1 = m1 ( g + a) = 3 ( 9.8 + 1.78) = 34.77 N

from equation (2)

T2 = m2 (g - a) = 4.5 (9.8 - 1.78) = 36.13 N

4 0

3 years ago

Hello!

34kurt

Iodine-131 has a half life of 8 days, so half of it is gone every 8 days.
10 grams of iodine-131 is left for 24 days.
At 8 days: 10/2=5 grams left
At 16 days: 5/2=2.5 grams left
At 24 days: 2.5/2=1.25 grams left.
**
Your mistake is that you stopped at 16 days.

8 0

3 years ago