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Maru [420]
3 years ago
13

The force diagram represents a girl pulling a sled with a mass of 6.0 kg to the left with a force of 10.0 N at a 30.0 degree ang

le. There is a 1.5 N force of friction to the right. The force of gravity is 58.8 N. What is the normal force acting on the sled? Round the answer to the nearest whole number. N What is acceleration of the sled? Round the answer to the nearest tenth. m/s2
Physics
2 answers:
Troyanec [42]3 years ago
7 0

Answer:

54N and -1.2m/s squared

Explanation:

STatiana [176]3 years ago
6 0

the correct answers are 54N and -1,2m/s^2

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Ariana is accelerating her car at a rate of 4.6 m/s2 for 10 seconds. Her starting velocity was 0 m/s.
Oksanka [162]

answer is 46 m/s

hope it help you

3 0
3 years ago
A freely suspended magnet does not point exactly at the geographical N-S direction​
elena-s [515]
Because their is nothing at the geographical poles that attracts the magnet
6 0
4 years ago
What answer is this? Pls help I’m freaking out
kolbaska11 [484]

Answer:

do u still need help it has been 2 weeks ??

Explanation:

8 0
3 years ago
Letting D D represent the maximum displacement, the extremes of the block's motion are at position A, where x = − D x=−D, and at
Ksju [112]

Answer:

The answer is at x = 0, which represents position B

Explanation:

The full question is:

"A block is attached to a horizontal spring and set in a

simple harmonic motion, as shown from above in the figure. When the spring is relaxed, the block is a position B, where the displacement x from the equilibrium position is 0. Letting D represent the maximum displacement, the extremes of the block's motion are at position A, where x= -D, and at position C, where x= D.

At what point in the motion is the speed of the block at its maximum?"

And you can see the figure on the attached file.

Simple Harmonic motion equations

We can start from the equation that describes the position that is

x(t)=D \sin\left(\omega t)

Here D stands for the amplitude which is the maximum displacement, and \omega is the angular velocity, thus we can find the derivative to find the velocity equation, so we get

v(t)=D \omega \cos (\omega t)

And we can find the derivative again to find the acceleration.

a(t) = -D\omega^2 \sin (\omega t)

Maximum speed

We reach the maximum speed when the acceleration equation is equal to 0,

a(t) =0\\-D\omega^2 \sin (\omega t)=0

Thus it happens when

\sin (\omega t)=0

So if we replace that on the position equation we get

x(t)=D \sin(\omega t) \\x(t)=D(0)\\x(t)=0

Thus the position where the speed of the block is at at its maximum is when it is going back to the origin, that is x = 0, so point b.

7 0
3 years ago
Water flows into a swimming pool at the rate of 5.85 gal/min. If the pool dimensions are 21.2 ft wide, 46.1 ft long and 19.4 ft
Lana71 [14]

First, calculate volume:

volume = 21.2 ft * 46.1 ft * 19.4 ft

volume = 18,960 ft^3

 

Convert to gallon:

volume = 18,960 ft^3 * (12 in / 1 ft)^3 * (1 gallon / 231 in^3)

volume = 141,830.71 gallon

 

Therefore the time is:

time = 141,830.71 gallon / (5.85 gal/min)

time = 24,244.57 min

6 0
3 years ago
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