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3 years ago

Beth moves a 15 N book 20 meters in 10 seconds. How much power was produced?

Physics

1 answer:

Mice21 [21]3 years ago

5 0

Answer:

30 Watts

Explanation:

Power = Work/Time

Work = Force*Distance

Power = Force * Distance / Time

Power = 15 N * 20 meters / 10 sec

Power = 30 Watts

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A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil

netineya [11]

$vi^{2}sin2thita/g =12^{2}sin2[30]/9.8=12.7$ Answer:

Explanation:

range is given as

6 0

3 years ago

Read 2 more answers

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0σC is fired at 21.0 i^ m/s straight toward a second

MissTica

(a)

Determine the system's initial configuration at ri = infinite particle separation and the system's final configuration at the point of closest approach.

Since the two-particle system is not being affected by any outside forces, we may treat it as an isolated system for momentum and use the momentum conservation law.

m1v1 + m1v2 = (m1+m2)v

The second particle's starting velocity is zero, so:

m1v1 = (m1+m2)v

After substituting the values we get,

v = 6i m/s

(b)

Since the two particle system is also energy-isolated, we may use the energy-conservation principle.

dK + dU = 0

Ki +Ui = Kf + Uf

Substituting the values,

1/2m1v1^2i + 1/2 m2v2^2i + 0 = 1/2m1v1^2f + 1/2m2v2^2f +ke q1q2/rf

The second particle's initial speed is 0 (v2 = 0). Additionally, both the first and second particle's final velocity have the same value, v. Put these values in place of the preceding expression:

1/2m1v1^2i = 1/2m1v1^2 + 1/2m2v2^2 +ke q1q2/rf

After solving we get,

rf = 2ke q1q2 / m1v1^2 - (m1+m2)v^2

Substituting the values we get,

rf = 3.64m

(c)

v1f = (m1-m2 / m1 + m2) v1i

v1f = -9i m/s

(d)

v2f = (2m1/ m1 +m2) v1i

After substituting the values,

v2f = 12i m/ s

Question :

Review. From a large distance away, a particle of mass 2.00 g and charge 15.0 \muμC is fired at 21.0 m/s straight toward a second particle, originally stationary but free to move, with mass 5.00 g and charge 8.50 \muμC. Both particles are constrained to move only along the x axis. (a) At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity. (b) Find the distance of closest approach. After the interaction, the particles will move far apart again. At this time, find the velocity of (c) the 2.00-g particle and (d) the 5.00-g particle. \hat{i}

To learn more about momentum conservation law click on the link below:

brainly.com/question/7538238

#SPJ4

5 0

2 years ago

What factor affects weight( what force causes weight to change)

Alex Ar [27]

Answer:

Gravitational force affects weight, weight changes with change in gravity

Hope it helped u,

mark as the brainliest

^_^

3 0

3 years ago

You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j

dmitriy555 [2]

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

= $mg=20\times 9.8 =196\ N$

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

7 0

3 years ago

The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 3 months. The line

Sladkaya [172]

Answer:

Explanation:

given

T = 3months = 7.9 × 10⁶s

orbital speed = 88 × 10³m/s

V= 2πr÷T

∴ r = (V×T) ÷ 2π

r = (88km × 7.9 × 10⁶s) ÷ 2π

r = 1.10 × 10⁸km

using kepler's 3rd law

mass of both stars = (seperation diatance)³/(orbital speed)²

M₁ + M₂ = (2r)³/( $\frac{1}{4}$ year)²

= (1.06 × 10²⁵)/(6.2×10¹³)

1.71×10¹²kg

since M₁ = M₂ =1.71×10¹²kg ÷ 2

M₁ = M₂ = 8.55×10¹¹kg

6 0

3 years ago