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3 years ago

What is the overall moment?

Physics

1 answer:

Setler79 [48]3 years ago

3 0

A moment causes a rotation about or axis. If the moment is to be taken about a point due to a force F, then in order for a moment to develop, the line of action cannot pass through that point...... the total moment was zero because the moment arm was zero as well

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A large crate sits on the floor of a warehouse. Paul and Bob apply constant horizontal forces to the crate. The force applied by

Delicious77 [7]

Answer:

W = -510.98J

Explanation:

Force = 43N, 61° SW

Displacement = 12m, 22° NE

Work done is given as:

W = F*d*cosA

where A = angle between force and displacement.

Angle between force and displacement, A = 61 + 90 + 22 = 172°

W = 43 * 12 * cos172

W = -510.98J

The negative sign shows that the work done is in the opposite direction of the force applied to it.

6 0

3 years ago

Which two labeled points on the fig are out of phase by 180° ?

GREYUIT [131]

You can use the points of answer:

B and F

5 0

3 years ago

During a collision between a photon and an electron, there is conservation of

jasenka [17]

The collision between a photon or a bundle of energy and an electron is known as the Compton Effect in which there is a transference of energy and momentum by the photon to the recoiling electron. It must be noted that both energy and momentum are being conserved during this elastic collision. After such phenomena, the photon acquires energy (represented by hf/) and the electron also has acquired a kinetic energy (represented by K).

7 0

4 years ago

Describe what is happening to the speed during the period (I). 0s - 10s __________________________________________________ (II).

aleksley [76]

Answer:

- There was a constant acceleration at 0 to 10s

- There was a zero acceleration at 10 to 25s

- There was a constant deceleration at 25 to 30s

Explanation:

See attachment for complete question.

Solving (a): What happens at 0s to 10s

There was a constant acceleration and this is proven below.

At time 0, velocity = 15

At time 10, velocity = 30

This is represented as:

$(t_1,v_1) = (0,15)$

$(t_2,v_2) = (10,30)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{30-15}{10 - 0}$

$A = \frac{15}{10}$

$A = 1.5$

Since the acceleration is positive, then it shows a constant acceleration.

Solving (b): What happens at 10s to 25s

There was a zero acceleration and this is because the velocity do not change.

See proof below

At time 10, velocity = 30

At time 25, velocity = 30

This is represented as:

$(t_1,v_1) = (10,30)$

$(t_2,v_2) = (25,30)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{30-30}{25 - 10}$

$A = \frac{0}{15}$

$A = 0$

Solving (c): What happens at 25s to 30s

There was a constant deceleration and this is proven below.

At time 25, velocity = 30

At time 30, velocity = 0

This is represented as:

$(t_1,v_1) = (25,30)$

$(t_2,v_2) = (30,0)$

Acceleration (A) is the rate of change of velocity against time.

So:

$A = \frac{v_2 - v_1}{t_2-t_1}$

$A = \frac{0-30}{30-25}$

$A = \frac{-30}{5}$

$A = -6$

Since the acceleration is negative, then it shows a constant deceleration

4 0

3 years ago

Magnitude of u = 15, direction angle θ = 35° Magnitude of v = 18, direction angle θ = 60° Find the magnitude and direction angle

djverab [1.8K]

Answer:

32.225 and angle is 48.7 degree.

Explanation:

u = 15, θ = 35 degree

v = 18, θ = 60

First represent the u and v in vector form.

u = 15 (Cos 35 i + Sin 35 j ) = 12.287 i + 8.6 j

v = 18 ( Cos 60 i + Sin 60 j ) = 9 i + 15.6 j

The sum of the two vectors is given by

u + v = 12.287 i + 8.6 j + 9 i + 15.6 j = 21.28 i + 24.2 j

Magnitude of u + v = $\sqrt{21.28^{2}+24.2^{2}}$ = 32.225

Let Ф be the angle

tan Ф = 24.2 / 21.28

Ф = 48.7 degree

5 0

3 years ago