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3 years ago

How long can a person dribble a ball when being closely guarded by a

Physics

1 answer:

Katena32 [7]3 years ago

7 0

Answer:

four seconds

Explanation:

A player who catches the ball and is being closely guarded may hold the ball for four seconds, then dribble the ball for four seconds, then hold the ball for another four seconds, then pass, will not be in violation of this rule.

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If 175 kJ of heat caused a chunk of ice at 0°C to melt to liquid water at 0°C, how many moles were in the ice? How many grams of

elena55 [62]

Answer is 29.1 mol or 525 g

Explanation:

8 0

3 years ago

An 80-kg astronaut becomes separated from his spaceship. He is 15.0 m away from it and at rest relative to it. In an effort to g

9966 [12]

The astronaut will take 300 seconds

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, the total momentum of the astronaut+object system must be conserved.

Initially, they are both at rest, so their total momentum is zero:

$p=0$

After the astronaut throws the object, their total momentum is:

$p=MV+mv$

where:

M = 80 kg is the mass of the astronaut

V is the final velocity of the astronaut

m = 500 g = 0.5 kg is the mass of the object

v = 8.0 m/s is the velocity of the object

Since momentum is conserved, we can write

$0=MV+mv$

And solving for V,

$V=-\frac{mv}{M}=-\frac{(0.5)(8.0)}{80}=-0.05 m/s$

Which means that he starts moving at 0.05 m/s in the direction opposite to the object.

Now the astronaut needs to cover a distance of

d = 15.0 m

And his speed is

v = 0.05 m/s

Therefore, the time taken is

$t=\frac{d}{v}=\frac{15.0}{0.05}=300 s$

Learn more about momentum here:

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#LearnwithBrainly

3 0

3 years ago

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the mete

disa [49]

Answer:

Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.

(a) Determine the position of wire 3.

b) Determine the magnitude and direction of current in wire 3

Explanation:

a) $F_{net} \text {on wire }3=0$

$\frac{\mu_0 I_1 I_3}{2 \pi x} = \frac{\mu I_2 I_3}{2 \pi (0.2+x)} \\\\\frac{1.5}{x} =\frac{4}{0.2+x} \\\\0.03+1.5x=4x\\\\x=0.012m\\\\=1.2cm$

position of wire = 50 - 1.2

= 48.8cm

b) $F_{net} \text {on wire }1=0$

$\frac{\mu _0 I_1 I_3}{2 \pi (1.2)} = \frac{\mu _0 I_1 I_2}{2 \pi (20)} \\\\\frac{I_3}{1.2} =\frac{4}{20} \\\\I_3=0.24A$

Direction ⇒ downward

5 0

3 years ago

Suppose astronomers built a 20-meter telescope. How much greater would its light-collecting area be than that of the 10-meter Ke

nirvana33 [79]

Answer:

4 times greater

Explanation:

Step 1: Calculate light-collecting area of a 20-meter telescope (A₁) by using area of a circle.

Area of circle = π*r² = $\frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 20-m

$A_{1} = \frac{\pi d^{2}}{4}$

$A_{1} = \frac{\pi (20^{2})}{4}$

A₁ = 314.2 m²

Step 2: Calculate light-collecting area of a 10-meter Keck telescope (A₂)

$A_{2} = \frac{\pi d^{2}}{4}$

Where d is the diameter of the circle = 10-m

$A_{2} = \frac{\pi (10^{2})}{4}$

A₂ = 78.55 m²

Step 3: divide A₁ by A₂

$= \frac{314.2 m^2}{78.55 m^2}$

= 4

Therefor, the 20-meter telescope light-collecting area would be 4 times greater than that of the 10-meter Keck telescope.

5 0

3 years ago

A centrifuge is used to test space pilots. The centrifuge spins with a centripetal acceleration of 3.04g. If the length of the a

uranmaximum [27]

Answer:

$approx.= 25\frac{m}{s}$

Explanation:

Centripetal acceleration (a) is defined as the square of an object's velocity (V^2) divided by the distance of the object from it's point/axis of revolution (r). So:

$a=\frac{V^{2} }{r}$

which allows us to solve for the velocity:

$V=\sqrt{ar}\\ Since: a=3.04g=(3.04)(9.81),r=21;\\V=\sqrt{(3.04)(9.81)(21)} =25.02...$

3 0

3 years ago