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mestny [16]
2 years ago
10

PLEASE HELP

Physics
2 answers:
Oksana_A [137]2 years ago
5 0

Answer:

The answer is A

Explanation: hope this helps :)

noname [10]2 years ago
3 0

Answer:

D

Explanation:

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Help with 5-10 please
Zina [86]

Answer:

explained in attachment

Explanation:

explained in attachment

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5 0
3 years ago
A 0.50-kg block slides across a tabletop with an initial velocity of 20 cm/s and comes to rest in a distance of 70 cm. Find the
mars1129 [50]

Answer:

FAE= 0.014 N

Explanation:

The KE of block is decreased because of the slowing action of the friction force .

Change in KE of block = work done on block by friction ƒ

⠀ ➪ ½mu²ƒ - ½mu²i = Fƒs cos θ

Because the friction force on the block is opposite in direction to the displacement , cos θ = -1

➢ Using Uƒ = 0 , Vƒ = 0.20 m/s , and s = 0.70 m

✒ We find ,

➪½mu²ƒ - ½mu²i = Fƒs cos θ

➪0-½ (0.50 kg) (0.20 m/s)² = (Fƒ) (0.70 m) (-1)

➪ Fƒ = 0.014 N

Hope this helped, can i pls have brainliest

6 0
3 years ago
Read 2 more answers
Doc Brown holds on to the end of the minute hand of the clock atop city hall. The tangential velocity of the minute hand is 0.41
alexdok [17]

The Professor's centripetal acceleration is 0.044 m/s²

Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.

It is given by:

a = v²/r

where v is the velocity and r is the radius.

Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:

a = v²/r = 0.419²/4 =  0.044 m/s²

The Professor's centripetal acceleration is 0.044 m/s²

Find out more at: brainly.com/question/6082363

3 0
2 years ago
A long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire. The wire has a charge
Alexxx [7]

Explanation:

It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.

The charge per unit length of the wire is \lambda and the net charge per unit length is 2 \lambda.

We know that there exist zero electric field inside the metal cylinder.

(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let \lambda_i\ and\ \lambda_o are the charge per unit length on the inner and outer surfaces of the cylinder.

For inner surface,

\phi=\dfrac{q_{enclosed}}{\epsilon_o}

E.A=\dfrac{q_{enclosed}}{\epsilon_o}

0=\dfrac{\lambda_i+\lambda}{\epsilon_o}

\lambda_i=-\lambda  

For outer surface,

\lambda_i+\lambda_o=2\lambda

-\lambda+\lambda_o=2\lambda

\lambda_o=3\lambda

(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :

E_o=\dfrac{\lambda_o}{2\pi \epsilon_o r}

E_o=\dfrac{3\lambda}{2\pi \epsilon_o r}

Hence, this is the required solution.

6 0
3 years ago
What did newton's law of universal gravitation tell us about how gravity works?
vazorg [7]
The force of gravity form the Sun will be stronger on a n object with more mass
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