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Simora [160]
3 years ago
13

Two objects exert a gravitational force of 3 N on one another when they are 10 m apart. What would that force be if the distance

between the two objects were reduced to 5 m?
A. 1.5 N
B. 6 N
C. 12 N
D. 24 N
Physics
1 answer:
jeka943 years ago
8 0
Gravitational force = G ( m1 m2 ) / r²
3 = G ( m1 m2 ) / ( 10 )²x = G ( m1 m2 ) / ( 5 )²We shall divide those two equations:3 / x = 1/100 / 1/25 = 25 / 100 = 1 / 4x · 1 = 3 · 4x = 12Answer:C. 12 N
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How much work is done on an object that is moved to acquire a displacement of 5 meters when 500 Newtons of force was exerted?​
grandymaker [24]

Answer:

2,500 Joules (J) or Newton Meter (N M)

Explanation:

Work = Force x Distance

The force in this equation is 500 Newtons. The distance (displacement) is 5 meters. Plug it into the equation above.

Work = 5m x 500n

Work = 2,500 Joules or Newton-Meters.

Therefore 2,500 Joules or Newton Meters of work is done on an object.

3 0
2 years ago
The Pacific Giant Kelp grows at a rate of 18 in/day. How many centimeters per hour is this?
Airida [17]

For this case we must do a conversion. By definition we have to:

1 inch equals 2.54 centimeters

1 day equals 24 hours.

So, we have:18 \frac {in} {day} * \frac {1} {24} \frac {day} {h} * \frac {2.54} {1} \frac {cm} {in} =

Canceling units we have:

\frac {18 * 1 * 2.54} {24 * 1} \frac {cm} {h} =\\\frac {45.72} {24}\frac {cm} {h} =

1,905 \frac {cm} {h}

Answer:

1,905 \frac {cm} {h}

5 0
3 years ago
A resistor R and another resistor 2R are connected in a series across a battery. If heat is produced at a rate of 10W in R, then
alekssr [168]

Answer:

Option B

Solution:

As per the question:

Heat produced at the rate of 10 W  

The resistor R and 2R are in series.

Also, in series, same current, I' passes through each element in the circuit.

Therefore, current is constant in series.

Also,

Power,P' = I'^{2}R

When current, I' is constant, then

P' ∝ R

Thus

\frac{P'}{2R} = \frac{10}{R}

P' = 20 W

5 0
3 years ago
A lighthouse is located on a small island 5 km away from the nearest point p on a straight shoreline and its light makes two rev
lisabon 2012 [21]

The distance starting from the point to the lighthouse would be regarded as the hypotenuse.

And also will be the radius of the circle the beam of light is generating at that point. 


So get the radius first

r = sqrt (1^2 + 5^2)

r = 5.099 km


find the circumference:

C = 2*pi * 5.099 km

C = 2 * 16.01898094

C = 32.04 km


Then find the speed in km/sec

One revolution: 60/2 = 30 sec per revolution

Speed = 32.04 km/30 sec

S = 1.068 km/sec is the speed of light

7 0
3 years ago
You are working as a summer intern for the Illinois Department of Natural Resources (DNR) and are assigned to some initial work
hammer [34]

Answer:

1)   F = 71.6 N , 2)    F = 120.2 N , 3)  P_m=  68.600 Pa, 4)  V = 2.4210-5 m³

Explanation:

This is a problem of fluid mechanics, to find the force we must use its definition

         P = F / A

         F = P A

The area of ​​the circular pipe is

         A = π r² = π d²/4

The pressure is given by the expression

         P = P_atm + ρ g h

1) the force on the outer side is

       P = P_atm

we substitute in the expression of force

         F = P_atm π d² / 4

         

let's calculate

         F = 1,013 10⁵ π 0.03²/4

         F = 7.16 10¹ N

         F = 71.6 N

2) the force on the inner side

  the pressure

       P = P_atm + ρ g h

       P = 1,015 10⁵ + 1,000 9.8 7

        P = 1,701 10⁵ Pa

        F = 1,701 10⁵ π 0.03² / 4

        F = 1,202 10²

        F = 120.2 N

3) manometric pressure

       Pm = ρ g h

       P m = 1000 9.8  7

       P_m=  68.600 Pa

4) In this part they ask for the volume that comes out in time t= 3 h

   to calculate this volume we can use the flow ratio

         Q = A v

          V t = A v

          V = A v / t

sent is the velocity of the water that comes out, to calculate it we use the Bernoulli equation

we will use index 1 for the lake surface and ionice 2 apa the position of the plug

           P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂

As the lake has much more capacity than the pipeline, the velocity of the surface of the lake is peeling, in this case we approximate it steel

           (P₁-P₂) + ρ G (y₁ -y₂) = ρ g v₂²

           1000 9.8 v₂² = ρ g h + 1000 9.8 (7-0)

             9800 v₂² = 1000 9.8 7 + 68600

              v₂ = √ (137200)

               v₂ = 370.4 m / s

             t = 3 h (3600s / h) = 10800 s

     

we substitute in the volume equation

             V = π d²/4   370.4 / 10800

             V = 2.4210-5 m³

4 0
2 years ago
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