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CaHeK987 [17]
2 years ago
7

The distance of the earth from the sun is 93 000 000 miles. if there are 3.15 × 107 s in one year, find the speed of the earth i

n its orbit about the sun.
Physics
2 answers:
faltersainse [42]2 years ago
5 0

The angular velocity of the orbit about the sun is:

w = 1 rev / year = 1 rev / 3.15 × 10^7 s

 

Now in 1 rev there is 360° or 2π rad, therefore:

w = 2π rad / 3.15 × 10^7 s

 

To convert in linear velocity, multiply the rad /s by the radius:

v = (2π rad / 3.15 × 10^7 s) * 93,000,000 miles

<span>v = 18.55 miles / s = 29.85 km / s</span>

Naily [24]2 years ago
3 0

Answer:

The speed of the earth about the sun is 4730.15 m/s.

Explanation:

It is given that,

The distance of the earth from the sun is, d=93000000\ miles=1.49\times 10^{11}\ m

Time taken to move from the earth to the sun, t=3.15\times 10^7\ s

We need to find the speed of the earth in its orbit about the sun. Let it is given by v. The total distance covered divided by total time taken is called the speed of the earth. Mathematically, it is given by :

v=\dfrac{d}{t}

v=\dfrac{1.49\times 10^{11}\ m}{3.15\times 10^7\ s}

v = 4730.15 m/s

So, the speed of the earth in its orbit about the sun is 4730.15 m/s. Hence, this is the required solution.

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A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w
Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0
1 year ago
A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
2 years ago
During the construction of an office building, a hammer is accidentally dropped from a height of 784 ft. the distance (in feet)
Serga [27]
T= 24.5 feet per second. That is the velocity it reaches at the end of its fall
7 0
2 years ago
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A rabbit is trying to cross the street. Its velocity v as a function of time t is given in the graph below where
Amiraneli [1.4K]

Answer:

2.5

Explanation:

8 0
2 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
Lady bird [3.3K]

Answer:

The options are approximations of the exact answers:

A) 1\times10^6N/C

B) 2\times10^{-13}N

C) 1\times10^{-2}V

D) Toward the inner wall

E) 3\times10^{-17}J

Explanation:

A) The electric field in a parallel plate capacitor is given by the formula E=\frac{\sigma}{k\epsilon_0}, where \epsilon_0=8.85\times10^{-12}C/Vm and in our case \sigma=10^5C/m^2 and, for air,k=1.00059, so we have:

E=\frac{10^5C/m^2}{(1.00059)(8.85\times10^{-12}C/Vm)}=1.13\times10^6N/C

B) The K+ ion has one elemental charge excess, so its charge is q=1.6\times10^{-19}C, and the force a charge experiments under an electric field E is given by F=qE, so we have:

F=(1.6\times10^{-19}C)(1.13\times10^6N/C)=1.81\times10^{-13}N

C) The potential difference between two points separated a distance d under an uniform electric potential E is given by \Delta V=dE, so we have:

\Delta V=(10\times10^{-9}m)(1.13\times10^6N/C)=1.13\times10^{-2}V

D) The electic field goes from positive to negative charges, so it goes towards the inner wall.

E) The work done by an electric field through a potential difference \Delta V on a charge Q is W=Q\Delta V, and is equal to the kinetic energy imparted on it, so we have:

K=(3\times10^{-15}C)(1.13\times10^{-2}V)=3.39\times10^{-17}J

5 0
2 years ago
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