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kondaur [170]
3 years ago
11

What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?​

Physics
1 answer:
givi [52]3 years ago
6 0

Answer:

1.2 x 10^-12

Explanation:

3/2.5 x 10^-6/10^6

1.2 x 10^-6 x 10^-6

1.2 x 10^-12

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A clock radio is rated as 30 w of power output. if the radio also draws 30 w at 120 v, which will the current draw be?
g100num [7]
P = IV

I = P/V =  30 / 120 = 0.25 A.

Current = 0.25A  
4 0
3 years ago
ALL OF MY POINTS FOR THIS!
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Newton's 2nd law of motion:

                       Force  =  (mass) x (acceleration)

                                   =  (1,127 kg) x (6 m/s² forward)

                                   =  (1,127 x 6)  newtons forward

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             Momentum  =  (mass) x (speed)

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3 0
4 years ago
A 200.0 kg piano is elevated by a crane to a height of 10.0 meters above a sidewalk. If the rope holding the piano breaks, what
Nitella [24]
All that business about the crane and the rope and the falling
is only there to confuse us.

The piano ended up 5 meters above the ground.

           Potential energy = (mass) (gravity) (height)

                                   = (200 kg) (9.81 m/s²) (5 m)

                                   = (200 · 9.81 · 5)  (kg-m²/s²)

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5 0
3 years ago
Consider an object rolling off of your lab bench. Discuss how you might be able to make measure- ments to determine its initial
guapka [62]

Answer:

measure the position every so often with a stopwatch

Explanation:

A possible method of measurement is to place a measuring tape along the path and measure the position every so often with a stopwatch, with this we can make a graph of position against time and by extrapolation find the initial velocity.

This is a method used in measurements of uniform movements of bodies

7 0
3 years ago
A 3.9 g dart is fired into a block of wood with a mass of 24.6 g. The wood block is initially at rest on a 1.5 m tall post. Afte
Galina-37 [17]

Answer:

46.48m/s

Explanation:

The problem is a combination of the principle of conservation of linear momentum and projectile motion.

The principle of conservation of linear momentum states that in a closed system, the total momentum of colliding bodies before impact is equal to the total momentum after impact. The masses stated in the problem experienced an inelastic collision. In an inelastic collision, the bodies involved stick together after the collision and move with a common velocity.

For two bodies of masses m_1 and m_2 moving with velocities u_1 and u_2 before impact, if they experience inelastic collision, the conservation of their momenta is as stated in equation (1);

m_1u_1+m_2u_2=(m_1+m_2)v..................(1)

were v is their common velocity after impact. If the second mass m_2 was at rest before the impact, then its initial velocity u_2=0m/s. therefore m_2u_2=0. Equation (1) then becomes;

m_1u_1=(m_1+m_2)v..............(2)

In the problem stated, the second mass taken as the mass of the wooden block was at rest before the impact and the collision was inelastic since both the wood and the dart stuck together and moved with a common velocity after the impact. Therefore we can use equation (2) for the problem.

Given;

m_1=3.9g=0.0039kg\\u_1=?\\m_2=24.6g=0.0246kg\\v=?

Substituting these values into (2), we get the following;

0.0039*u_1=(0.0039+0.024)v\\0.0039u_1=0.0285v.........(3)

Their common v velocity after impact now makes both the wooden block and the dart (as a single body) to fall vertically through a height h of 1.5m over a range R of 3.5m as stated by the problem; hence by the principle of projectile motion for a body projected horizontally, the following relationship holds;

R= vt............(4)

were t is the time taken to fall through the height h. To obtain t we use the second equation of free fall under gravity;

h=\frac{1}{2}gt^2...........(5)

were g is acceleration due to gravity taken as 9.8m/s^2. Therefore;

1.5=\frac{1}{2}*9.8*t^2\\1.5=4.9t^2\\t^2=\frac{1.5}{4.9}=0.306\\t=\sqrt{0.306} =0.55s

We then substitute R and t into equation (4) to obtain v.

3.5=v*0.55\\v=\frac{3.5}{0.55}\\v=6.36m/s

We now further substitute this value of v into (3) to obtain u_1;

u_1=\frac{0.0285v}{0.0039}\\\\u_1=\frac{0.0285*6.36}{0.0039}\\\\u_1=\frac{0.18126}{0.0039}\\\\u_1=46.48m/s

4 0
4 years ago
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