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Alex73 [517]
2 years ago
11

suppose a ball is thrown vertically upward. Eight seconds later it returns to its point of release. What is the initial velocity

of the ball?
Physics
1 answer:
valentinak56 [21]2 years ago
6 0
The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.

At its highest point, its velocity changed from upward to downward. 
At that instant, its velocity was zero.

The acceleration of gravity is 9.8 m/s².  That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second. 

-- If the object is falling downward, it moves 9.8 m/s faster every second.

-- If the object is tossed upward, it moves 9.8 m/s slower every second.

The ball took 4 seconds to lose all of its upward speed.  So it must have
been thrown upward at  (4 x 9.8 m/s)  =  39.2 m/s .

(That's about  87.7 mph straight up.  Somebody had an amazing pitching arm.)
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Why the bulb of thermometer in cylindrical
crimeas [40]
I'm not accurately sure if you're asking for why the bulb of a thermometer is in a cylindrical shape. So let me continue. The shape of the which is thin and cylindrical in the shape is the increase of the effect of mercury in the tube to rise and fall depending on the contact temperature.  
4 0
3 years ago
Car accelerates 86km/hr in 4.6s . What is the average acceleration in m/s^2
never [62]

speed = 86km/hr

= 86000/3600

= 23.889m/s

time = 4.6s

acceleration = 23.889/4.6

a = 5.19m/s^2

4 0
3 years ago
three carts of masses 2 kg, 18 kg, and 9 kg move on a frictionless horizontal track with speeds of 10m/s, 8m/s, and 2m/s. the ca
Elan Coil [88]

Answer:

V = 6.3 m/s

Explanation:

Given:

m₁ = 2 kg

m₂ = 18 kg

m₃ = 9 kg

V₁ = 10 m/s

V₂ = 8 m/s

V₃ = 2 m/s

__________

V - ?

Let us write the momentum conservation law for an inelastic impact:

m₁·V₁ + m₂·V₂ + m₃·V₃ = (m₁ +m₂ + m₃) ·V

Cart speed after interaction:

V = ( m₁·V₁ + m₂·V₂ + m₃·V₃ ) / (m₁ +m₂ + m₃)

V = (2·10 + 18·8 + 9·2) / ( 2 + 18 + 9) = 182 / 29 ≈  6.3 m/s

4 0
1 year ago
A car rounds a curve. The radius of curvature of the road is R, the banking angle with respect to the horizontal is θ and the co
exis [7]

Answer:

v = √[gR (sin θ - μcos θ)]

Explanation:

The free body diagram for the car is presented in the attached image to this answer.

The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.

Resolving the weight into the axis parallel and perpendicular to the inclined plane,

N = mg cos θ

And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ

Frictional force = Fr = μN = μmg cos θ

Centripetal force responsible for keeping the car in circular motion = (mv²/R)

So, a force balance in the plane parallel to the inclined plane shows that

Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)

(mv²/R) = (mg sin θ - μmg cos θ)

v² = R(g sin θ - μg cos θ)

v² = gR (sin θ - μcos θ)

v = √[gR (sin θ - μcos θ)]

Hope this Helps!!!

5 0
2 years ago
Which is greater 63cm or 6m​
slavikrds [6]

Answer:

6m is greater

Explanation:

1 meter is equal to 100 cm and 63cm is less than 1m

3 0
3 years ago
Read 2 more answers
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