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kumpel [21]
2 years ago
6

The 0.5kg soccer ball moves toward the net with a force of 4N. What is its acceleration?​

Physics
1 answer:
Eduardwww [97]2 years ago
6 0

Answer:

8 m/s²

Explanation:

Given,

Force ( F ) = 4 N

Mass ( m ) = 0.5 kg

To find : -

Acceleration ( a ) = ?

Formula : -

F = ma

a = F / m

= 4 / 0.5

= 40 / 5

a = 8 m/s²

It's acceleration is 8 m/s².

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7 0
3 years ago
When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed sh
aev [14]

Answer:

Explanation:

Given

Initial speed is u=V

Maximum height of Pebble is H

Deriving maximum height of Pebble and considering motion in vertical direction

v^2-u^2=2 as

where v=final velocity

u=initial velocity

a=acceleration

s=Displacement

Final velocity will be zero at maximum height

0-(V)^2=2\times (-g)\cdot H

H=\frac{V^2}{2g}

i.e. maximum height is dependent on square of initial velocity

for twice the height

2H=\frac{(V')^2}{2g}

on comparing

V'=\sqrt{2}V

7 0
3 years ago
A motorcycle of mass 250 kg drives around a circle with a centripetal acceleration of 3.4 m/s2. What is the centripetal force ac
Vlad [161]
Hope this helps and have a nice day!

8 0
2 years ago
Read 2 more answers
Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×10^4)
anyanavicka [17]

Answer:

1.4 billion light years away

Explanation:

v = Recessional velocity = 30000 km/s[/tex]

H_0 = Hubble constant = \frac{65}{3.2\times 10^6}\ ly

D = Distance to the galaxy

According to Hubble's law

v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly

The galaxy is 1.4 billion light years away

5 0
2 years ago
A test tube of length L and cross-sectional area A is submerged in water with the open end down so that the edge of the tube is
Margaret [11]

Answer:

if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

Explanation:

The air in the tube can be considered an ideal gas,

           P V = nR T

In that case we have the tube in the air where the pressure is P1 = P_atm, then we introduce the tube to the water to a depth H

For pressure the open end of the tube is

         P₂ = P_atm + ρ g H

Let's write the gas equation for the colon

            P₁ V₁ = P₂ V₂

            P_atm V₁ = (P_atm + ρ g H) V₂

             V₂ = V₁    P_atm / (P_atm + ρ g h)

If the air obeys Boyle's law e; volume within the had must decrease due to the increase in pressure, if we measure the change in height of the gas within the had and obtain a straight line in relation to the depth we can conclude that the air complies with Boye's law.

The main assumption is that the temperature during the experiment does not change

6 0
2 years ago
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