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2 years ago

11

Two bowling balls each have a mass of 6.8 kg. They are located next to each other with their centers 21.8 cm apart. What gravita

tional force do they exert on each other?

1 answer:

12345 [234]2 years ago

3 0

Answer:

6.49 x 10^-8 N

Explanation:

formula is

F= G * ((m1 * m2)/r^2)

F = 6.67x10^-11 * ((6.8*6.8/.218)

F = 6.49 x 10^-8 Newtons

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What is the jet stream and how does it separate cold air to warm air?

docker41 [41]

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(credit to google)

4 0

3 years ago

A rod of length Lo moves iwth a speed v along the horizontal direction. The rod makes an angle of (θ)0 with respect to the x' ax

Colt1911 [192]

Answer:

From the question we are told that

The length of the rod is $L_o$

The speed is v

The angle made by the rod is $\theta$

Generally the x-component of the rod's length is

$L_x = L_o cos (\theta )$

Generally the length of the rod along the x-axis as seen by the observer, is mathematically defined by the theory of relativity as

$L_xo = L_x \sqrt{1 - \frac{v^2}{c^2} }$

=> $L_xo = [L_o cos (\theta )] \sqrt{1 - \frac{v^2}{c^2} }$

Generally the y-component of the rods length is mathematically represented as

$L_y = L_o sin (\theta)$

Generally the length of the rod along the y-axis as seen by the observer, is also equivalent to the actual length of the rod along the y-axis i.e $L_y$

Generally the resultant length of the rod as seen by the observer is mathematically represented as

$L_r = \sqrt{ L_{xo} ^2 + L_y^2}$

=> $L_r = \sqrt{[ (L_o cos(\theta) [\sqrt{1 - \frac{v^2}{c^2} }\ \ ]^2+ L_o sin(\theta )^2)}$

=> $L_r= \sqrt{ (L_o cos(\theta)^2 * [ \sqrt{1 - \frac{v^2}{c^2} } ]^2 + (L_o sin(\theta))^2}$

=> $L_r = \sqrt{(L_o cos(\theta) ^2 [1 - \frac{v^2}{c^2} ] +(L_o sin(\theta))^2}$

=> $L_r = \sqrt{L_o^2 * cos^2(\theta) [1 - \frac{v^2 }{c^2} ]+ L_o^2 * sin(\theta)^2}$

=> $L_r = \sqrt{ [cos^2\theta +sin^2\theta ]- \frac{v^2 }{c^2}cos^2 \theta }$

=> $L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Hence the length of the rod as measured by a stationary observer is

$L_r = L_o \sqrt{1 - \frac{v^2}{c^2 } cos^2(\theta ) }$

Generally the angle made is mathematically represented

$tan(\theta) = \frac{L_y}{L_x}$

=> $tan {\theta } = \frac{L_o sin(\theta )}{ (L_o cos(\theta ))\sqrt{ 1 -\frac{v^2}{c^2} } }$

=> $tan(\theta ) = \frac{tan\theta}{\sqrt{1 - \frac{v^2}{c^2} } }$

Explanation:

7 0

3 years ago

A Nichrome wire 44 cm long and 0.30 mm in diameter is connected to a 3.1 V flashlight battery. What is the electric field inside

Alexeev081 [22]

Answer:

7.05 Volts/m

Explanation:

L = length of the Nichrome wire = 44 cm = 0.44 m

V = Potential difference across the end of the wire = battery voltage = 3.1 Volts

E = magnitude of electric field inside the wire

Magnitude of electric field inside the wire is given as

$E = \frac{V}{L}$

Inserting the values

$E = \frac{3.1}{0.44}$

E = 7.05 Volts/m

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2 years ago

A rowboat that is traveling 2.5m/s to the north, encounters a current traveling 4.3m/s to the east. What is the resultant veloci

saveliy_v [14]

Pablo clavó un clavito

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2 years ago

A 500 kg object is hanging from a spring attached to the ceiling. If the spring constant in the spring is 900 N/kg, how far does

My name is Ann [436]

We are given that a 500 kg object is hanging from a spring. To determine the amount the spring is stretched we will use Hook's law, which states the following:

$F=kx$

Where:

$\begin{gathered} F=\text{ force} \\ k=\text{ spring constant} \\ x=\text{ distance stretched} \end{gathered}$

Since the object is hanging the only force acting on the spring is the weight of the object. The weight of the object is:

$F_g=mg$

Where:

$\begin{gathered} m=\text{ mass} \\ g=\text{ acceleration of gravity} \end{gathered}$

Plugging in the values we get:

$F_g=(500\operatorname{kg})(9.8\frac{m}{s^2})$

Solving the operations:

$F_g=4900N$

Now we solve for "x" from Hook's law by dividing both sides by "k":

$\frac{F}{k}=x$

Now we plug in the known values:

$\frac{4900N}{900\frac{N}{m}}=x$

Solving the operations:

$5.4m=x$

Therefore, the spring is stretched by 5.4 meters.

7 0

7 months ago