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brilliants [131]
3 years ago
8

Time (min)

Physics
1 answer:
charle [14.2K]3 years ago
8 0

Answer:

rgvdghrryyrfgtr was the other monkey that I have to do with my best pet peeve for the rest and neon scooter is my friend is giving her the money for your cuz to be a good offer tho it was a great idea for a while and then she said yes and she is a great person ever and she was madly happy I have to go soon to be honest with me if she wants me back to my base or not but she wasnt it to be honest with her but

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Need help ??? Please
Yanka [14]
Is there information in the previous question which relates to this one?
8 0
3 years ago
A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of
grigory [225]

Answer:

a=2.9\ m/sec^2

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force F_r acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

\vec F_r=m\vec a

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force F_r is computed as the hypotenuse of a right triangle

|F_r|=\sqrt{256^2+104^2}

|F_r|=276.32\ Nw

The acceleration can be obtained from the formula

F_r=ma

Note we are using only magnitudes here

\displaystyle a=\frac{F_r}{m}

\displaystyle a=\frac{276.32Nw}{95.3Kg}

\boxed{a=2.9\ m/sec^2}

7 0
3 years ago
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
2 years ago
Mechanical locks can accept a variety of inputs as keys, including magnetic strips on ID cards, radio signals from name badges,
lidiya [134]

Answer:

The answer is B) False

3 0
3 years ago
An orange might roll of your cafeteria try when you stop suddenly because of
guapka [62]

Answer:

Explanation:

Newton's first law of motion states that an object in motion stays in motion. The orange is moving and then the tray stops making the orange move forward because of inertia.

3 0
2 years ago
Read 2 more answers
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