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zvonat [6]
2 years ago
6

a student balances a 1.5kg of broom by placing her finger 1.4m from the end of the broom handle. How far from the broom handle w

ould she have to balance the broom if she hung a 400g mass from the end of the broom handle?​
Physics
1 answer:
Delicious77 [7]2 years ago
4 0

The broom handle that she have to balance if she hung a 400g mass from the end of the broom handle is 5.24m

This problem is centered on moment. Moment is the turning effect of a force about a point. It is expressed as:

Moment = Force× Distance

According to principle of moment, the sum of clockwise moment is equal to sum of anticlockwise moment at shown

M1d1 = M2d2

Given the following

M1 = 1.5kg

d1 = 1.4m

M2 = 400g = 0.4kg

d2 is required

Substitute

1.5(1.4) = 0.4d2

2.1 = 0.4d2

d2 = 2.1/0.4

d2 = 5.24m

Hence the broom handle that she have to  if she hung a 400g mass from the end of the broom handle is 5.24m

Learn more here: brainly.com/question/21945515

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The pages of a book are numbered 1 to 200 and each
never [62]

Answer:

10.4mm

Explanation:

2 pages = 1 leaf

200 pages = 100 leaves

100 × 0.10 = 10 mm thickness

Total thickness = 2(0.20) +10 = 0.4+10 = 10.4mm

6 0
3 years ago
In an elastic collision, a 580 kg bumper car collides directly from behind with a second, identical bumper car that is traveling
kozerog [31]

Answer:  vl = 2.75 m/s vt = 1.5 m/s

Explanation:

If we assume that no external forces act during the collision, total momentum must be conserved.

If both cars are identical and also the drivers have the same mass, we can write the following:

m (vi1 + vi2) = m (vf1 + vf2) (1)

The sum of the initial speeds must be equal to the sum of the final ones.

If we are told that kinetic energy must be conserved also, simplifying, we can write:

vi1² + vi2² = vf1² + vf2² (2)

The only condition that satisfies  (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:

vf1 = vi2 and vf2 = vi1

If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:

vf1 = 2.75 m/s  vf2 = 1.5 m/s

8 0
3 years ago
Calculate the change in velocity of a 0.070kg tennis ball hit by Serena with a force of 140 N over 0.020 s
Brilliant_brown [7]
V=at and a=F/m

140/.070 = 2000m/s^2

2000*.020 = 40m/s

The ball’s velocity increased by 40m/s.
6 0
3 years ago
(I) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second pro
Kipish [7]

Answer:

The second projectile was 1.41 times faster than the first.

Explanation:

In the ballistic pendulum experiment, the speed (v) of the projectile is given by:  

v = \frac{m + M}{m} \cdot \sqrt{2gh}

<em>where m: is the mass of the projectile, M: is the mass of the pendulum, g: is the gravitational constant and h: is the maximum height of the pendulum.   </em>

To know how many times faster was the second projectile than the first, we need to take the ratio for the velocities for the projectiles 2 and 1:    

\frac{v_{2}}{v_{1}} = \frac{\frac{m_{2} + M}{m_{2}} \cdot \sqrt{2gh_{2}}}{\frac{m_{1} + M}{m_{1}} \cdot \sqrt{2gh_{1}}}           (1)

<em>where m₁ and m₂ are the masses of the projectiles 1 and 2, respectively, and h₁ and h₂ are the maximum height reached by the pendulum by the projectiles 1 and 2, respectively.  </em>

Since the projectile 1 has the same mass that the projectile 2, we can simplify equation (1):

\frac{v_{2}}{v_{1}} = \frac{\sqrt{h_{2}}}{\sqrt{h_{1}}}  

\frac{v_{2}}{v_{1}} = \frac{\sqrt{5.2 cm}}{\sqrt{2.6 cm}}

\frac{v_{2}}{v_{1}} = 1.41  

Therefore, the second projectile was 1.41 times faster than the first.  

I hope it helps you!

8 0
3 years ago
An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane
lubasha [3.4K]

Answer:

\rho = 12580.7 kg/m^3

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

F = \frac{GMm}{(r + h)^2}

here we have

F =\frac {mv^2}{(r+ h)}

\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}

here we have

v = \sqrt{\frac{GM}{(r + h)}}

now we can find time period as

T = \frac{2\pi (r + h)}{v}

T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}

1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}

M = 4.54 \times 10^{23} kg

Now the density is given as

\rho = \frac{M}{\frac{4}{3}\pi r^3}

\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}

\rho = 12580.7 kg/m^3

8 0
2 years ago
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