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mylen [45]
3 years ago
15

Which objects will likely have the greatest gravitational force between them?

Physics
1 answer:
Mariulka [41]3 years ago
3 0

Answer:

d. Two soccer balls that are touching each other

Explanation:

Let m_1 be the mass of a tennis ball, m_2 is the mass of a soccer ball.

As the mass of a soccer ball is more than the mass of a tennis ball, so

m_2 > m_1

Let d_1 be the distance between the centers of both the balls near each other and d_2 be the distance between the centers of both the balls touching each other.

So, d_2 > d_1

The gravitational force, F, between the two objects having masses M and m and separated by distance d is

F=\frac{GMm}{d^2}

Where G is the universal gravitational constant.

As, the gravitational force is directly proportional to the product of both the masses and inversely proportional to the square of the distance between them,  so selecting the larger mass (m_2, soccer ball) separated by a lesser distance (d_2, touching) to get more gravitational force.

Therefore, there will be a larger gravitational force between them when two soccer balls touching each other.

Hence, option (d) is correct.

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Aleksandr-060686 [28]
Water and h20 delivery have been a very meaningful company to the community
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ASK YOUR TEACHER A baseball with a mass of 146 g is thrown horizontally with a speed of 40.6 m/s (91 mi/h) at a bat. The ball is
RideAnS [48]

Answer:

Explanation:

mass of the ball = 146 g = 146 / 1000 = 0.146 kg

initial speed of the ball = 40.6 m/s

final speed of the ball = - 45.1 m/s

time of impact = 1.05 ms = 1.05 / 1000 = 0.00105 s

impulse, Ft = change in momentum = mv - mu = m (v-u)

F = m (v - u) / t = 0.146 kg ( -45.1 -40.6) / 0.00105 s = -11916.4 N

4 0
3 years ago
List materials that are not good conductors of energy
snow_lady [41]
Almost anything that isn't a metal. Rubber, pure water, wood, and plastic, are all good answers.
7 0
4 years ago
Read 2 more answers
Running at 1.55 m/s, Bruce, the 40.0 kg quarterback, collides with Biff, the 90.0 kg tackle, who is traveling at 7.0 m/s in the
Margarita [4]

Answer:Bruce is knocked backwards at  

14

m

s

.

Explanation:

This is a problem of momentum (

→

p

) conservation, where

→

p

=

m

→

v

and because momentum is always conserved, in a collision:

→

p

f

=

→

p

i

We are given that  

m

1

=

45

k

g

,  

v

1

=

2

m

s

,  

m

2

=

90

k

g

, and  

v

2

=

7

m

s

The momentum of Bruce (

m

1

) before the collision is given by

→

p

1

=

m

1

v

1

→

p

1

=

(

45

k

g

)

(

2

m

s

)

→

p

1

=

90

k

g

m

s

Similarly, the momentum of Biff (

m

2

) before the collision is given by

→

p

2

=

(

90

k

g

)

(

7

m

s

)

=

630

k

g

m

s

The total linear momentum before the collision is the sum of the momentums of each of the football players.

→

P

=

→

p

t

o

t

=

∑

→

p

→

P

i

=

→

p

1

+

→

p

2

→

P

i

=

90

k

g

m

s

+

630

k

g

m

s

=

720

k

g

m

s

Because momentum is conserved, we know that given a momentum of  

720

k

g

m

s

before the collision, the momentum after the collision will also be  

720

k

g

m

s

. We are given the final velocity of Biff (

v

2

=

1

m

s

) and asked to find the final velocity of Bruce.

→

P

f

=

→

p

1

f

+

→

p

2

f

→

P

f

=

m

1

v

1

f

+

m

2

v

2

f

Solve for  

v

1

:

v

1

f

=

→

P

f

−

m

2

v

2

f

m

1

Using our known values:

v

1

f

=

720

k

g

m

s

−

(

90

k

g

)

(

1

m

s

)

45

k

g

v

1

f

=

14

m

s

∴

Bruce is knocked backwards at  

14

m

s

.

Explanation:

5 0
3 years ago
A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature
Fofino [41]

Answer:

Density change, Δρ = 2.4 × 10⁻⁴ kg/m³

Temperature Change, ΔT = 0.0258 K

Velocity Change, Δc = 0.0148 m/s

Explanation:

For sound waves moving through the air,

Pressure and Temperature varies thus

(P₀/P) = (T₀/T)^(k/(k-1))

Where P₀ = initial pressure of air = 101KPa = 101000 Pa

P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa

T₀ = initial temperature of air = 30°C = 303.15 K

T = final temperature of air = ?

k = ratio of specific heats = Cp/Cv = 1.4

(101000/101030) = (303.15/T)^(1.4/(1.4-1))

0.9990703 =(303.15/T)^(3.5)

Solving This,

T = 303.1758 K

ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K

Density can be calculate in two ways,

First method

Δρ = ρ - ρ₀

P₀ = ρ₀RT₀

ρ₀ = P₀/RT₀

R = gas constant for air = 287 J/kg.k

where all of these are values for air before the wave propagates

P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K

ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³

ρ = P/RT

P = 101030 Pa, T = 303.1758K

ρ = 101030/(287×303.1758) = 1.1611115 kg/m³

Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³

Second method

(ρ₀/ρ) = (T₀/T)^(1/(k-1))

Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.

Velocity Change

c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s

c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s

Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s

QED!

5 0
3 years ago
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