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Leni [432]
3 years ago
10

Prior to determining the experimental design, a scientist typically? A. makes observations. B. forms a hypothesis. C. performs a

n experiment. D. predicts the result of an experiment.
Physics
1 answer:
Umnica [9.8K]3 years ago
7 0

Prior to determining the experimental design, a scientist typically forms a hypothesis. The answer is letter B. this is to prepare the scientist, the possible outcome of their research before the experimental design whether they are wrong or not.

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A 2.0 kg stone is tied to a 0.30 m string and swung around a circle at a constant angular velocity of 12.0 rad/s. The net torque
saul85 [17]

Answer: τ = 0

Explanation:

At constant angular velocity there is no angular acceleration therefore no torque.

τ = Iα

6 0
3 years ago
A rock exerts 5000 Pa of pressure on the ground. If the rock weighs 250 N, how much area is in contact with the ground?
ololo11 [35]

P = F/S - S = F/P = 250/5000= 0.05 m2

4 0
3 years ago
Read 2 more answers
A racing car on a straight accelerates from 100mph to 316 mph in three seconds what is the acceleration?
omeli [17]

Answer:

72mph/sec

Explanation:

The car goes from 100mph to 316mph in three seconds. Meaning it increases its speed by (316 - 100)mph in three seconds. That is 216 mph increase in three seconds. So, we divide the speed increase by the amount of time the increase occurred over. We get:

216mph / 3sec = 72mph/sec, our final answer

Hope it made sense. I would appreciate Brainliest, but no worries.

8 0
3 years ago
Suppose we are given a square coil that is 5.5 cm on a side containing 100 loops of very fine wire. The total resistance of this
maw [93]

Answer:

I = 578A

Explanation:

The magnitude of the peak value of the induced current flowing in a coild is given by

I = \frac{\epsilon_{max}}{R}

I = \frac{NBA\omega}{R}

Where

I= current

N = Number of loops

\omega = angular velocity

R= Resistance

B = Magnetic field

A = Area

Replacing our values we have that,

I = \frac{(100)(1.3)(0.055)^2(2.5)}{1.7*10^{-3}}

I = 578A

8 0
3 years ago
Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t
elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play .  But the horizontal range covered by the body thrown

depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

Horizontal distance covered = t x horizontal velocity

If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .

If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t  + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only  horizontal velocity initially , Vy = 0

h = 1/2 gt²

t = \sqrt{\frac{2h}{g} }

Horizontal distance covered  = t x Vx

= \sqrt{\frac{2h}{g} } \times  V_x

From this expression also

Horizontal distance covered is proportional to \sqrt{h} .

7 0
3 years ago
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