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Daniel [21]
3 years ago
7

Assapp!!!’!!!!!!!! !!!!!!

Physics
1 answer:
dlinn [17]3 years ago
6 0

Answer:

delta r(x) = (delta (r)) * cos(alpha), delta r(y) = (delta(r)) * sin(alpha)

Explanation:

Well it's a simple rule I guess...

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A baseball (m=140g) traveling 32m/s moves a fielders
posledela

Answer:

Average force, F = 286.72 N

Explanation:

Given that,

Mass of the baseball, m = 140 g = 0.14 kg

Speed of the ball, v = 32 m/s

Distance, h = 25 cm = 0.25 m

We need to find the average  force exerted by the ball on the glove. It is solved using the conservation of energy as :

\dfrac{1}{2}mv^2=mgh

F = mg

\dfrac{1}{2}mv^2=Fh

F=\dfrac{mv^2}{2h}

F=\dfrac{0.14\times (32)^2}{2\times 0.25}

F = 286.72 N

So, the average force exerted by the ball on the glove is 286.72 N. Hence, this is the required solution.

7 0
3 years ago
A 2.0 kg block is released from rest at the top of a curved incline in the shape of a quarter of a circle of radius R = 3.0 m. T
Zigmanuir [339]

The block has maximum kinetic energy at the bottom of the curved incline. Since its radius is 3.0 m, this is also the block's starting height. Find the block's potential energy <em>PE</em> :

<em>PE</em> = <em>m g h</em>

<em>PE</em> = (2.0 kg) (9.8 m/s²) (3.0 m)

<em>PE</em> = 58.8 J

Energy is conserved throughout the block's descent, so that <em>PE</em> at the top of the curve is equal to kinetic energy <em>KE</em> at the bottom. Solve for the velocity <em>v</em> :

<em>PE</em> = <em>KE</em>

58.8 J = 1/2 <em>m v</em> ²

117.6 J = (2.0 kg) <em>v</em> ²

<em>v</em> = √((117.6 J) / (2.0 kg))

<em>v</em> ≈ 7.668 m/s ≈ 7.7 m/s

3 0
3 years ago
Review. An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT . The angular momen
Ghella [55]

The speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

The angular momentum(L) of an electron moving in a circular path is given by the formula,

L = mvr ........(i)

We know that the radius of the path of an electron in a magnetic field is

r = mv/qB

Putting this value in equation (i),

L = mv x mv/qB

or L = (mv)^2/qB

Putting the given values in the above equation,

4 x 10^-25 = (9.1x10^-31)^2 x v^2/ 1.6 x 10^-19 x 1 x 10^-3

v comes out to be 8.88 x 10^7 m/s.

Hence, the speed of an electron when it moves in a circular path perpendicular to a constant magnetic field is 8.88 x 10^7 m/s.

To know more about "angular momentum", refer to the following link:

brainly.com/question/15104254?referrer=searchResults

#SPJ4

5 0
1 year ago
Find the weight of an object of mass 5 kg
Elis [28]

Answer:

weight on earth is mg

which is 5*9.8

49 Newton

weight on moon is 1/6 th of weight on earth

1/6*49

8.166 Newton..

3 0
2 years ago
Read 2 more answers
On a perfect fall day, you are hovering at low altitude in a hot-air balloon, accelerated neither upward nor downward. The total
Stella [2.4K]

Explanation:

Since the balloon is not accelerating means that the net force on the balloon is zero. This implies that the weight of balloon must be equal to the buoyant force on balloon.

Hence, the buoyant force equals the weight of air displaced by the balloon, also 20,000 N.

Weight of the air displaced = density of air × volume

The density of air at 1 atm pressure and 20º C is 1.2 kg/m³  

the volume V = 20,000/(1.2×9.8) =  1700 m³

3 0
3 years ago
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