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Amiraneli [1.4K]
3 years ago
6

Darryl throws a basketball at the gym floor. The ball bounces once on the floor and comes to rest in his coach’s hands. At which

point are all the forces on the basketball balanced?
Physics
1 answer:
mina [271]3 years ago
8 0

Answer:

when the ball is at rest in his coach's hands.

Explanation:

The forces on the basketball are balanced when the basketball is not experiencing any acceleration. This happens when the ball is in his coach's hand: in fact, at that moment the ball is at rest, so it means that its acceleration is zero. According to Newton's second law, this also mean that the net force on the basketball is zero, so the forces on the ball are balanced:

F=ma

where F is the net force, m is the mass of the ball and a is the acceleration.

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Star A and Star B have measured stellar parallax of 1.0 arc second and 0.75 arc second, respectively. Which star is closer? How
zhuklara [117]

Answer:

Star A is closer than Star B

Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

Here we can say

angle = \frac{1 Parsec}{distance}

so we have

distance = \frac{1 Parsec}{angle}

so here we have

angle subtended by Star A = 1 arc sec

angle subtended by star B = 0.75 arc sec

now we have

distance for star A is given as

d_a = \frac{1 Parsec}{1} = 1 Parsec

distance of star B is given as

d_b = \frac{1 Parsec}{0.75} = 1.33 Parsec

So star A is closer than star B

7 0
3 years ago
Numbered 8 nowwwwwww
amm1812

Torque =  r x F

|F| =  mg =  60 * 10 N = 600 N ( assuming g ~ 10m/s^2)

distance of fulcrum = torque / Force = 90/600 m = .15 m.

7 0
3 years ago
When does The velocity of a wave change
dmitriy555 [2]

Answer:

hvy hbv evuheuhvjhfv

Explanation:

vb vbb  fdvywibvhqieiv

6 0
3 years ago
A bungee jumper of mass m jumps off a bridge. Assume that the bungee cord behaves like am ideal spring of spring constant k. Whe
DanielleElmas [232]

Answer:

b) √[(kx²/m) - 2gx]

Explanation:

The energy at the lowest point is equal to:

E_{elas}=\frac{1}{2} *k*x^{2}

where:

Eelas = elastic energy [J]

k = spring constant [N/m]

x = extension of the spring [m]

We consider the lowest point, as the point where the potential energy is zero. At the moment when the person goes back through the point of the normal length of the elastic cord, it is this point that the person will have potential energy and kinetic energy.

E_{elas}=E_{pot}+E_{kine}\

\frac{1}{2}*k*x^{2}=m*g*x +\frac{1}{2} *m*v^{2}  \\v^{2} = \frac{k*x^{2} }{m}-2*g*x\\ v=\sqrt{\frac{k*x^{2} }{m}-2*g*x}

8 0
3 years ago
If an object of mass 70kg falls from a height of 500 m, what is the maximum velocity of the object?
Umnica [9.8K]

Answer:

H = 1/2 * g * t^2     since initial velocity is zero

v = g * t    where v is the final velocity

t = v / g

H = 1/2 g * v^2 / g^2 = 1/2 v^2 / g

v = (2 * H * g)^1.2

v = (2 * 500 * 9.8)^1/2 = 99 m/s

Check: t = v / g = 99 / 9.8 = 10.1 sec

H = 1/2 * 9.8 * 10.1^2 = 500 m

7 0
3 years ago
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