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3 years ago

How long will it take for an object to hit the ground when dropped from a height of 7m?

Physics

1 answer:

White raven [17]3 years ago

3 0

Answer:

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s.

Explanation:

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The sidereal period of the moon around the Earth is 27.3 days. Suppose a satellite were placed in Earth orbit, halfway between E

icang [17]

Answer: 9.7 days

Explanation:

Applying Kepler's 3rd law, we can write the following proportion:

(Tm)² / (dem)³ = (Tsat)² / (dem/2)³

(As the satellite is placed in an orbit halfway between Erth's center and the moon's orbit).

Simplifyng common terms, and solving for Tsat, we have:

Tsat = √((27.3)²/8) = 9.7 days

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2 years ago

What are the formulae of momentun and their time of use

Ivanshal [37]

Answer:

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6 0

2 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

Now, gravitational force on particle 3 due to particle 1:

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

gravitational force on particle 3 due to particle 2:

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

Now the net force

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

For angle in counterclockwise direction from the +x-axis

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

4 0

3 years ago

A large solar panel on a spacecraft in Earth orbit produces 1.0 kW of power when the panel is turned toward the sun. What power

Mandarinka [93]

Answer:

e*P_s = 11 W

Explanation:

Given:

- e*P = 1.0 KW

- r_s = 9.5*r_e

- e is the efficiency of the panels

Find:

What power would the solar cell produce if the spacecraft were in orbit around Saturn

Solution:

- We use the relation between the intensity I and distance of light:

I_1 / I_2 = ( r_2 / r_1 ) ^2

- The intensity of sun light at Saturn's orbit can be expressed as:

I_s = I_e * ( r_e / r_s ) ^2

I_s = ( 1.0 KW / e*a) * ( 1 / 9.5 )^2

I_s = 11 W / e*a

- We know that P = I*a, hence we have:

P_s = I_s*a

P_s = 11 W / e

Hence, e*P_s = 11 W

3 0

2 years ago

When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a fo

Sladkaya [172]

Answer:

The answers to the questions are;

The force the biceps must exert to hold a 850 g ball at the end of the forearm at distance dball = 34.0 cm from the elbow, with the forearm parallel to the floor is 166.77 N Upwards.

The force the elbow must exert is 145.6785 N downwards.

Explanation:

We are required to analyze the system using the principle of moments as follows.

Mass of forearm = 1.30 kg

Weight of forearm = mass × acceleration due to gravity = 1.30 kg × 9.81 m/s²= 12.753 N

Location of point of action of weight of forearm = midpoint 17 cm

Forearm to biceps distance = 3.00 cm from the elbow.

Mass of the ball = 850 g

Weight of ball = mass × acceleration due to gravity = 0.850 kg × 9.81 = 8.3385 N

Position of the ball = end of the forearm

Therefore, taking moment about the elbow, we have

∑Mₓ = 0, we take clockwise moment to be positive, therefore

8.3385 N× 34.0 cm + 12.753 N× 17 cm - $F_{biceps}$ ×3.00 cm = 0

Therefore $F_{biceps}$ ×3.00 cm = 500.31 N cm

Therefore $F_{biceps}$ = (500.31 N·cm)/(3.00 cm) = 166.77 N Upwards

The force exerted by the elbow is given by,

Taking moment about the point of contact between the biceps and the forearm, we get

$F_{elbow}$ × 3.00 cm = 8.3385 N× 31.0 cm + 12.753 N× 14 cm

$F_{elbow}$ = 145.6785 N downwards.

4 0

2 years ago