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3 years ago

Ted lifts a 10N weight at a height of 1.5 m in 1

Physics

1 answer:

masha68 [24]3 years ago

5 0

Answer:

D. Ted expanded more power.

Explanation:

Given the following data;

For Ted.

Force = 10N

Height = 1.5m

Time = 1 seconds

To find Ted's power;

Power = workdone/time

But workdone = force * distance

Workdone = 10 * 1.5

Workdone = 15 Nm

Power = 15/1

Power = 15 Watts.

For Johnny.

Force = 10N

Height = 1.5m

Time = 2 seconds

To find Ted's power;

Power = workdone/time

But workdone = force * distance

Workdone = 10 * 1.5

Workdone = 15 Nm

Power = 15/2

Power = 7.5 Watts

Therefore, from the calculations we can deduce and conclude that Ted expanded more power.

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The specific heat of Aluminum is 0.9 J/g K. How much heat is lost as a 100 gram sample of Aluminum is dropped into a beaker of w

otez555 [7]

Energy=mass*SHC*temp change
=100*0.9*75
=6750J

6 0

4 years ago

An object of mass 300 g, moving with an initial velocity of 5.00i-3.20j m/s, collides with an sticks to an object of mass 400 g,

Alexus [3.1K]

Answer:

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

Explanation:

Mass of object 1 , m₁ = 300 g = 0.3 kg

Mass of object 2 , m₂ = 400 g = 0.4 kg

Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s

Initial velocity of object 2 , v₂ = 3.00j m/s

Mass of composite = 0.7 kg

We need to find final velocity of composite.

Here momentum is conserved.

Initial momentum = Final momentum

Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s

Final momentum = 0.7 x v = 0.7v kgm/s

Comparing

1.5 i + 0.24 j = 0.7v

v = 2.14 i + 0.34 j

Magnitude of velocity

$v=\sqrt{2.14^2+0.34^2}=2.17m/s$

Direction,

$\theta =tan^{-1}\left ( \frac{0.34}{2.14}\right )=9.03^0$

Velocity is 2.17 m/s at an angle of 9.03° above X-axis.

7 0

3 years ago

Will Give Brainlist <br> Check pic for question

Harrizon [31]

Ohms law = v= Ir

V= 0.02 x 4000 = 80v

5 0

3 years ago

Read 2 more answers

lace each example of a collision in correct category. bumper cars colliding man jumping in a cab mud sticking to car hat being s

Marrrta [24]

Answer:

bumper cars colliding- inelastic

man jumping in a cab- perfectly inelastic

mud sticking to car - perfectly inelastic

hat being sat on door being slammed- inelastic

ball bouncing- elastic

Explanation:

In a perfectly inelastic collision, the objects stick together after collision and move with a common velocity. Maximum kinetic energy is lost during such collision.

For an inelastic collision, kinetic energy is partly lost and the colliding objects move apart at different velocities. This is often encountered in real life situations.

For an elastic collision, both momentum and kinetic energy are conserved. The object rebounds with the same relative velocity with which it approached.

3 0

4 years ago

The radius of a lead atom is 175 pm. how many lead atoms would have to be laid side by side to span a distance of 6.11 mm?

podryga [215]

The number of atoms needed to be laid side by side to span a distance of 6.11 mm is 17,457,143.

<h3>What is Unit conversion?</h3>

Unit conversion is a way of converting some common units into another without changing their real value. for, example, 1 centimeter is equal to 10 mm, though the real measurement is still the same the units and numerical values have been changed.

Given that the radius of a lead atom is 175 picometers. Therefore, the radius of the lead atom in mm can be written as,

1 picometer = 10⁻⁹ millimeter

175 picometer = 175×10⁻⁹ millimeter

The radius of the lead atom = 175×10⁻⁹ millimeter

Since the atoms are needed to be arranged side by side, therefore, the diameter of one lead atom will join with the other, thus, the diameter of the lead atom is,

Diameter of lead atom = 2 × 175 × 10⁻⁹ millimeter = 350 ×10⁻⁹ millimeter

Now, let the number of lead atoms be represented by n. Therefore, the number of atoms needed to be laid side by side to span a distance of 6.11 mm is,

6.11 mm = Number of atoms × Diameter of atoms

6.11 mm = n × 350 ×10⁻⁹ mm

Solving for n,

n = 6.11 mm / 350 ×10⁻⁹ mm

n = 17,457,142.86 ≈ 17,457,143

Hence, the number of atoms needed to be laid side by side to span a distance of 6.11 mm is 17,457,143.

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7 0

2 years ago