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Lapatulllka [165]
2 years ago
7

When 999mm is added to 100m ______ is the result​

Physics
2 answers:
mart [117]2 years ago
6 0

Answer:

when 999mm added to 100m 101m is the result.

Explanation:

i think so

Amanda [17]2 years ago
4 0

Answer:

what,     100.999m

Explanation:

convert 999 mm into meters, which is 0.999m and add that to a 100 m and that will make the total 100.999 m

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A circular coil 17.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati
sergeinik [125]

Answer:

The torque in the coil is  4.9 × 10⁻⁵ N.m  

Explanation:

T = NIABsinθ

Where;

T is the  torque on the coil

N is the number of loops = 9

I is the current = 7.8 A

A is the area of the circular coil = ?

B is the Earth's magnetic field = 5.5 × 10⁻⁵ T

θ is the angle of inclination = 90 - 56 = 34°

Area of the circular coil is calculated as follows;

A = \frac{\pi d^2}{4} \\\\A = \frac{\pi 0.17^2}{4} =0.0227 m^2

T = 9 × 7.8 × 0.0227 × 5.5×10⁻⁵ × sin34°

T = 4.9 × 10⁻⁵ N.m

Therefore, the torque in the coil is  4.9 × 10⁻⁵ N.m

5 0
1 year ago
If two waves with identical crests and troughs meet, what is happening?
Nastasia [14]
<span>If two waves with identical crests and troughs meet, what is happening?
</span>C. Constructive interference is occurring. 
3 0
2 years ago
Read 2 more answers
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, start fraction, start text, m, end tex
love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

\Delta x = v_{i}.t + \frac{a}{2}.t^{2}

v_{i} is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

\Delta x = 5(6) - \frac{4}{2}.6^{2}

\Delta x = 30 - 2.36

\Delta x = - 42

Displacement of the boat for t=6.0s interval is \Delta x = - 42m, i.e., 42 m to the left.

8 0
2 years ago
A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha
Alecsey [184]

Answer:

<h2>9.375Nm</h2>

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given  F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ  = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

6 0
2 years ago
What is the absolute value of the horizontal force that each athlete exerts against the ground?
alexandr402 [8]
Refer to the diagram shown below.

When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.

At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.

Answer:  
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.

6 0
2 years ago
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