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matrenka [14]
3 years ago
9

A ball has mass of 140g. what is the force to accelerate the ball at 25m/s

Physics
2 answers:
adelina 88 [10]3 years ago
8 0

Hello there!

formula: F=ma


It would be 3.5 N

Fittoniya [83]3 years ago
6 0

We know, F = m * a

Here, m = 140 g = 0.140 Kg

a = 25 m/s2

It would be: F = 0.140 * 25 = 3.5 N

So your answer would be 3.5N


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Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0
3 years ago
Propose an experiment that could be used to demonstrate that friction is a contact force?
vesna_86 [32]
Maybe push or pull an object with a large amount of mass? you are force a (pushing through object) aka making contact. i hope i helped not good with physics :)
3 0
3 years ago
Read 2 more answers
The magnitude of the magnetic flux through the surface of a circular plate is 5.90 10-5 T · m2 when it is placed in a region of
valkas [14]

Answer:

The strength of the magnetic field is 3.5 x 10⁻³ T

Explanation:

Given;

magnitude of the magnetic flux , Φ = 5.90 x 10⁻⁵ T·m²

angle of inclination of the field, θ = 42.0°

radius of the circular plate, r = 8.50 cm = 0.085 m

Generally magnetic flux in a uniform magnetic field is given as;

Φ = BACosθ

where;

B is the strength of the magnetic field

A is the area of the circular plate

Area of the circular plate:

A = πr²

A = π (0.085)² = 0.0227 m²

The strength of the magnetic field:

B = Φ / ACosθ

B = ( 5.90 x 10⁻⁵) / ( 0.0227 x Cos42)

B = 3.5 x 10⁻³ T

Therefore, the strength of the magnetic field is 3.5 x 10⁻³ T

3 0
3 years ago
A bicycle wheel of radius 0.70 m is rolling without slipping on a horizontal surface with an angular speed of 2.0 rev/s when the
aleksandrvk [35]

Answer:

θ= 5 radian

Explanation:

Given data:

Radius r = 0.70 m

Initial angular speed ω_i = 2rev/s

Time t = 5 s

Final angular speed ω_f =0

so we have angular displacement

\theta= \frac{\omega_f-\omega_i}{2}\times t

putting values

\theta= \frac{0-2}{2}\times5 = 5 rad

8 0
3 years ago
If the force on a spring is 2 N and it stretched 0.5 m, what is the spring constant?
andriy [413]
<h2>It solved by the Hooke's law states F=kx</h2>

answer is

<h2>0.4n/m</h2>
6 0
3 years ago
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