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3 years ago

Acceleration is directly proportional to net force. What is the relationship between acceleration and mass?

Physics

1 answer:

guajiro [1.7K]3 years ago

4 0

Answer:

They are inversely related; As acceleration decreases, mass increases; As acceleration decreases, mass increases.

Explanation:

This is to keep the force the same.

F=2m•0.5a

F=0.5m•2a

Both forces have the same value, but they have acceleration and masses that are inversely proportional to eachother.

In real life, you can see it with a recoil gun. The bullet and the gun have equal but opposite forces, but the bullet accelerates faster than the gun due to its smaller mass, while the gun recoils at a slower acceleration due to its larger mass.

You might be interested in

What is the energy in joules of a photon with a frequency of 3.16e 12 s-1?

erica [24]

We have: Energy(E) = Planck's constant(h) × Frequency(∨)
Here, Planck's constant(h) = 6.626 × 10⁻³⁴ J/s
Frequency (∨) = 3.16 × 10¹² /s

Substitute the values into the expression:
E = (6.626 × 10⁻³⁴)(3.16 × 10¹²) J
E = 2.093 × 10⁻²¹ Joules

In short, Your Final answer would be 2.093 × 10⁻²¹ J

Hope this helps!

5 0

3 years ago

Read 2 more answers

A pendulum is raised to a height of 0.3m above its lowest point and released. What is the velocity of the pendulum at its lowest

enyata [817]

Answer:

v = 2,425 m / s

Explanation:

A simple pendulum has anergy stored at the highest point of the path and this energy is conserved throughout the movement.

highest point

Em₀ = U = m g y

lowest point

$Em_{f}$ = K = ½ m v²

Em₀ = Em_{f}

mg y = ½ m v²

v = √ 2gy

let's calculate

v = √ (2 9.8 0.3)

v = 2,425 m / s

3 0

3 years ago

A box of mass 1 kg is hung from a spring scale. The reading on the spring scale is approximately 10N. What would be the reading

8090 [49]

The answer will be 50N.
This is because the spring reads weight and weight is mass times acceleration due to gravity.5kg*10m/s2=50N

8 0

3 years ago

Read 2 more answers

The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z

sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So, 4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0

3 years ago

If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?

Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

$y=y_{o}+V_{oy}t-\frac{g}{2}t^{2}$ (1)

Where:

$y=0$ is the final height of the ball

$y_{o}=h$ is the initial height of the ball

$V_{oy}=V_{o}sin(0\°)=0$ is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)