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3 years ago

A woman throws a baseball straight up into the air. what is the baseball's acceleration at its highest point?

1 answer:

5 0

After the ball leaves the woman's hand, the only relevant force acting on the ball is the gravity, which produces an acceleration of a=g=9.81 m/s^2 on the baseball, directed downward.
Gravity acts on the ball for the whole duration of the motion, so the acceleration of the ball is $g=9.81 m/s^2$ at every point of the motion, and therefore also when the baseball is at its highest point.

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What momentum of a 50kilogram ice skater gliding across the ice at a speed of 5m/s?

Ivenika [448]

Momentum (P) = Mass (kg) * Velocity (m/s)

P = M * V
P = 50 * 5
P = 250

So momentum is 250 kgm/s

5 0

3 years ago

Read 2 more answers

A particle with a mass of 6.64 × 10–27 kg and a charge of +3.20 × 10–19 C is accelerated from rest through a potential differenc

blondinia [14]

Answer:

Explanation:

Given that,

Mass m = 6.64×10^-27kg

Charge q = 3.2×10^-19C

Potential difference V =2.45×10^6V

Magnetic field B =1.6T

The force in a magnetic field is given as Force = q•(V×B)

Since V and B are perpendicular i.e 90°

Force =q•V•BSin90

F=q•V•B

So we need to find the velocity

Then, K•E is equal to work done by charge I.e K•E=U

K•E =½mV²

K•E =½ ×6.64×10^-27 V²

K•E = 3.32×10^-27 V²

U = q•V

U = 3.2×10^-19 × 2.45×10^6

U =7.84×10^-13

Then, K•E = U

3.32×10^-27V² = 7.84×10^-13

V² = 7.84×10^-13 / 3.32×10^-27

V² = 2.36×10^14

V=√2.36×10^14

V = 1.537×10^7 m/s

So, applying this to force in magnetic field

F=q•V•B

F= 3.2×10^-19 × 1.537×10^7 ×1.6

F = 7.87×10^-12 N

6 0

3 years ago

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Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a

DaniilM [7]

Answer:

a) $k = 1343.6\,\frac{N}{m}$ , b) $l = 0.501\,m\,(50.1\,cm)$

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

$F \propto \Delta l$

In this case, the force is equal to the weight of the object:

$F = m\cdot g$

$F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )$

$F = 80.614\,N$

The spring constant is:

$k = \frac{F}{\Delta l}$

$k = \frac{80.614\,N}{0.408\,m-0.348\,m}$

$k = 1343.6\,\frac{N}{m}$

b) The length of the spring is:

$F = k\cdot (l-l_{o})$

$l = l_{o} + \frac{F}{k}$

$l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }$

$l = 0.501\,m\,(50.1\,cm)$

6 0

3 years ago

imagine riding a single-speed bicycle. why do you have to push harders on the pedals to start the bicycle moving than to keep it

Flura [38]

A large force is required to accelerate the mass of the bicycle and rider. Once the desired constant velocity is reached, a much smaller force is sufficient to overcome the ever-present frictional forces.

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3 years ago

A Rankine cycle with one closed feedwater heater with its drain cascaded backward has a water mass flow rate through the steam g

mamaluj [8]

Answer:Draw a T-s diagram for the ideal Rankine Cycle

Explanation:

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3 years ago