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2 years ago

The following is regarding Momentum Change and Force. help is needed please!

Physics

1 answer:

mixer [17]2 years ago

4 0

The distance travelled is 10 m and the velocity gained at the end of this time is 2 m/s.

<h3>Velocity of the object at the end of the time</h3>

F = mv/t

where;

m is mass of the object
v is velocity of the object
t is time

Ft = mv

v = Ft/m

v = (50 x 10)/250

v = 2 m/s

<h3>Distance traveled by the object</h3>

v² = u² + 2as

where;

u is initial velocity = 0

a is acceleration

a = F/m

a = 50 N/ 250 kg

a = 0.2 m/s²

v² = 0 + 2as

s = v²/2a

s = (2²)/(2 x 0.2)

s = 10 m

Thus, the distance travelled is 10 m and the velocity gained at the end of this time is 2 m/s.

Learn more about distance here: brainly.com/question/2854969

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An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be

lord [1]

Answer:

the quarterback will be moving back at speed of 0.080625 m/s

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0

3 years ago

1. What is meant by half-life?

Bezzdna [24]

1. The time for a radioactive sample to reduce to half of its original mass.
4. 10% of 232 is 23.2 times 6 equals 139.2
68.9 x6 equals 413.4
5. 0.37kg
6. 2000 years
7. 6.84 seconds

8 0

3 years ago

Calculate the radius of the orbit of a proton moving at 2.2x10^6 m/s in a magnetic field 0.7 T where v and B are perpendicular.

Juliette [100K]

Answer:

3.28 cm

Explanation:

To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:

r = $\frac{mv}{qB}$