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2 years ago

14

a horizontal force of 300 N is needed to push a crate along the floor at constant speed. What is the friction force on the crate

?

1 answer:

finlep [7]2 years ago

5 0

Answer:

Since a 100-N force is being used to push the crate in one direction, and there is a total net force of 0, the force of friction acting on the crate must also be 100 N. This frictional force points in the direction opposite of the push.

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How much work is done to move 2.10 μC of charge from the negative terminal to the positive terminal of a 3.50 V battery?

rjkz [21]

Answer:

6.3E-6

Explanation:

Workdone = V Q

4 0

3 years ago

Pattinson Food Suppliers has received a letter from a government agency regarding some complaints of wrong labeling of meat prod

Anuta_ua [19.1K]

Answer:

1. The United States Department of Agriculture’s Food Safety and Inspection Service (FSIS)

Explanation:

FSIS has primary responsibility for the regulation of food labeling

for meat and poultry products under the FMIA(Federal Meat Inspection Act) and the PPIA(Poultry Products Inspection Acts) and is also

authorized to regulate food labeling for exotic species of animals under the

Agricultural Marketing Act of 1946.

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2 years ago

A handful of professional skaters have taken a skateboard through an inverted loop in a full pipe. For a typical pipe with a dia

Bingel [31]

Answer

given,

diameter of the pipe is = (14 ft)4.27 m

minimum speed of the skater must have at very top = ?

At the topmost point of the pipe the normal force will be equal to zero.

F = mg

centripetal force acting on the skateboard

$F = \dfrac{mv^2}{r}$

equating both the force equation

$mg = \dfrac{mv^2}{r}$

$v = \sqrt{gr}$

r = d/2 = 14/ 2 = 7 ft

or

r = 4.27/2 = 2.135 m

g = 32 ft/s² or g = 9.8 m/s²

$v = \sqrt{32 \times 7}$

v = 14.96 ft/s

or

$v = \sqrt{9.8 \times 2.135}$

v = 4.57 m/s

5 0

3 years ago

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

where the relation between time (t) and period (T) is:

$\mathsf{t = \dfrac{T}{2}}$

T = 2t

T = 2(0.247)

T = 0.494 seconds

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

By making the spring constant k the subject of the formula:

$\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}$

$\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}$

$\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}$

$\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}$

$\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}$

$\mathbf{ k =8.25 \ N/m}$

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

6 0

2 years ago

Two diverging light rays, originating from the same point, have an angle of 5° between them. After the rays reflect from a plane

LuckyWell [14K]

As a reference, consider the line from the point perpendicular to the mirror.
That direction is called 'normal' to the mirror.

The ray on the right leaves the point traveling 5° to the right of the normal,
and leaves the mirror on a path that's 10° to the right of the normal.

The ray on the left leaves the point traveling 5° to the left of the normal,
and leaves the mirror on a path that's 10° to the left of the normal.

The angle between the two rays after they leave the mirror is 20° .

Frankly, Charlotte, if there were more than 5 points available for this answer,
I'd seriously consider giving you a drawing too.

3 0

2 years ago

Read 2 more answers