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2 years ago

What is the gravitational force between two bowling balls with 7kg mass each and 1.1m apart ?

Physics

1 answer:

Rufina [12.5K]2 years ago

8 0

Answer:

2.7010 * 10^-9 N

Explanation:

solution;

mass (m1) = 7kg

mass (m2) = 7kg

distance(d) = 1.1m

universal gravitational constant (G) = 6.67 * 10^-11

we know that,

gravitational force = (G*m1*m2)/d^2

= ((6.67 * 10^-11) *7 *7)/(1.1)^2

=(3.2683 *10^-9)/1.21

= 2.7010 * 10^-9

Gravitational force = 2.7010 * 10^-9 N

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Please help!!! A river has a constant current of 3 km per hour. If a motorboat, capable of maintaining a constant speed of 20km

PilotLPTM [1.2K]

Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have:

theta = arcsin(3/20) = approx. 8.63° 

The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. 

x = sqrt(20^2 - 3^2) 
= sqrt(400 - 9) 
= sqrt 391 

The boat's crossing time = 
0.5 km/(sqrt 391 km/hr) 
= (0.5/sqrt 391) hr 
= approx. 0.025 hr 
= approx. 91 seconds

4 0

3 years ago

Which material produces static charge when rubbed?

natita [175]

Many materials produce static charge

5 0

3 years ago

How much heat is required to heat 2 kg of water from 25°C to 40°C?

Dominik [7]

Answer:

126000 J

Explanation:

Applying,

Q = cm(t₂-t₁).................. Equation 1

Where Q = Amount of heat, c = specifc heat capacity of water, m = mass of water, t₁ = Initial temperature, t₂ = Final temperature.

From the question,

Given: m = 2 kg, t₁ = 25°C, t₂ = 40°C

Constant: c = 4200 J/kg.°C

Substitute these value into equation 1

Q = 2×4200(40-25)

Q = 2×4200×15

Q = 126000 J

5 0

3 years ago

A solenoid coil with 22 turns of wire is wound tightly around another coil with 330 turns. The inner solenoid is 21.0 cm long an

horsena [70]

Answer:

The average magnetic flux through each turn of the inner solenoid is $11.486\times10^{-8}\ Wb$

Explanation:

Given that,

Number of turns = 22 turns

Number of turns another coil = 330 turns

Length of solenoid = 21.0 cm

Diameter = 2.30 cm

Current in inner solenoid = 0.140 A

Rate = 1800 A/s

Suppose For this time, calculate the average magnetic flux through each turn of the inner solenoid

We need to calculate the magnetic flux

Using formula of magnetic flux

$\phi=BA$

$\phi=\dfrac{\mu_{0}N_{2}I}{l}\times\pi r^2$

Put the value into the formula

$\phi=\dfrac{4\pi\times10^{-7}\times330\times0.140}{21.0\times10^{-2}}\times\pi\times(\dfrac{2.30\times10^{-2}}{2})^2$

$\phi=11.486\times10^{-8}\ Wb$

Hence, The average magnetic flux through each turn of the inner solenoid is $11.486\times10^{-8}\ Wb$

7 0

3 years ago

A body with the inertial

Andrews [41]

Answer:

Explanation:

Hi there,

To get started, recall the kinematic equations from either a textbook, equation sheet, etc. Kinematic equations are used when acceleration is constant, as stated in the prompt.

Best way to use kinematic equations is to see which variable you are looking for, then which variable is unknown to you and is not needed for that equation.

a) average velocity

Takes the form of:

$v_a_v_g=\frac{d_t_o_t_a_l}{t}=\frac{v+v_0}{2}$ this is the literal definition of average velocity; initial plus final divided by 2.

We know total displacement and total time elapsed, so we will use the middle form of the equation:

$v_a_v_g=\frac{1640m}{40s}=41 \ m/s$

b) the final velocity

We can still use the average velocity formula, as the other two equations that include final velocity have acceleration variable which is unknown as of now.

Solve for final velocity:

$v=(2v_a_v_g)-v_o = 2(41 \ m/s) - (8 m/s) = 74 m/s\\$ this makes sense, since a velocity later in time is higher than a velocity earlier in time. It is increasing with increasing time because of acceleration.

c) the acceleration

There are two equations that can be used to solve this, but we will use the less time-consuming one, but both produce same answer:

$a = \frac{v-v_0}{t_t_o_t_a_l} = \frac{(74-8)m/s}{40s} =1.65 m/s^{2}$

Notice, change in velocity over change in time, and acceleration is constant. When acceleration is constant, it models a linear function, and acc. is just slope!

Study well and persevere. If you liked this solution, hit Thanks or give a rating!

thanks,

3 0

3 years ago