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Sphinxa [80]
3 years ago
11

A person kicks a ball off of a 50m high cliff with a speed of 10 m/s. How long will it take the ball to hit the ground? * 7 poin

ts 2.3 sec 6.3 sec 4.5 sec 3.2 sec
Physics
1 answer:
Musya8 [376]3 years ago
6 0

Presumably, the ball is kicked parallel to the ground below the cliff, so its altitude <em>y</em> at time <em>t</em> is

y(t)=50\,\mathrm m-\dfrac12gt^2

where <em>g</em> = 9.80 m/s^2 is the acceleration due to gravity.

The ball hits the ground when <em>y</em> = 0:

0 = 50\,\mathrm m-\dfrac12gt^2

t^2=\dfrac{100\,\mathrm m}g

t=\dfrac{10}{9.80}\,\mathrm s\approx\boxed{3.2\,\mathrm s}

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You are performing an experiment that requires the highest possible energy density in the interior of a very long solenoid. Whic
Alinara [238K]

Answer:

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

Explanation:

As we know that energy density depends on the strength of the magnetic field. The magnetic field strength depends on the no of turns of the solenoid and the current passing through it. The greater the number of turns per unit length, greater the current passing through it, more stronger the magnetic field is. As

B = μ₀nI

n = no of turns

I = current through the wire

So the right options are

b. increasing the number of turns per unit length on the solenoid

e. increasing the current in the solenoid

5 0
3 years ago
Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each othe
deff fn [24]

Answer:

\theta_2 - \theta_1 = 156.93 degree

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

so we have

y = A sin\omega t

y = \frac{A}{5}

so now we have

\frac{A}{5} = A sin\omega t

now we have

\theta_1 = 11.53 degree

so the phase other particle in opposite direction is given as

\theta_2 = 180 - 11.53 = 168.46 degree

so we have phase difference given as

\theta_2 - \theta_1 = 168.46 - 11.53

\theta_2 - \theta_1 = 156.93 degree

7 0
3 years ago
1. A block is pulled to the right at constant velocity by a 20N force acting at 30o above the horizontal. If the coefficient of
DiKsa [7]

Answer:

44.6 N

Explanation:

Draw a free body diagram of the block.  There are four forces on the block:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

and applied force F pulling right 30° above horizontal.

Sum of forces in the y direction:

∑F = ma

N + F sin 30° − mg = 0

N = mg − F sin 30°

Sum of forces in the x direction:

∑F = ma

F cos 30° − Nμ = 0

F cos 30° = Nμ

N = F cos 30° / μ

Substitute:

mg − F sin 30° = F cos 30° / μ

mg = F sin 30° + (F cos 30° / μ)

Plug in values:

mg = 20 N sin 30° + (20 N cos 30° / 0.5)

mg = 44.6 N

8 0
2 years ago
Read 2 more answers
A 90. 0-kg ice hockey player hits a 0. 150-kg puck, giving the puck a velocity of 45. 0 m/s. If both are initially at rest and i
Mice21 [21]

The distance traveled by the hockey player is 0.025 m.

<h3>The principle of conservation of linear momentum;</h3>
  • The principle of conservation of linear momentum states that, the total momentum of an isolated system is always conserved.

The final velocity of the hockey play is calculated by applying the principle of conservation of linear momentum;

m_1v_1 = m_2 v_2\\\\v_1 = \frac{m_2 v_2}{m_1} \\\\v_1 = \frac{0.150 \times 45}{90} \\\\v_1 = 0.075 \ m/s

The time taken for the puck to reach 15 m is calculated as follows;

t = \frac{d}{v} \\\\t = \frac{15\ m}{45 \ m/s} \\\\t = 0.33 \ s

The distance traveled by the hockey player at the calculated time is;

d = vt\\\\d = 0.075 \ m/s \ \times 0.33 \ s\\\\d = 0.025 \ m

Learn more about conservation of linear momentum here: brainly.com/question/7538238

4 0
2 years ago
The basic barometer can be used as an altitude-measuring device in airplanes. The ground control reports a barometric reading of
kipiarov [429]

Answer:

Δh_air=714m

Explanation:

Given data

P_{1}=753mmHg\\P_{2}=690mmHg\\ p_{air}=1.2kg/m^{3}\\  g=9.8m/s^{2}

Solution

ΔP=P₁-P₂

=(ΔhHg)×pHg×g

=(Δh_air)× p_air ×g

Then

Δh_air=(pHg+ΔhHg)÷p_air

=\frac{13600*(753-690)*10^{-3} }{1.2}\\ =714m

Δh_air=714m

7 0
3 years ago
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