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Brut [27]
3 years ago
12

With what magnitude of force does a ball of mass 0.75 kilograms need to be hit so that it accelerates at the rate of 25 meters/s

econd2? Assume that the ball undergoes motion along a straight line
Physics
2 answers:
serious [3.7K]3 years ago
8 0

     Force = (mass) x (acceleration)

               = (0.75 kg) x (25 m/s²)

               =  (0.75 x 25)  kg-m/s²

               =    18.75 newtons .

Note that even though we're talking about a 'hit', the acceleration only
lasts as long as the bat is in contact with the ball.  Once the ball leaves
the bat, it travels at whatever speed it had at the instant when they parted. 
Any change in its speed or direction after that is the result of gravity, air
resistance, and the fielder's mitt.  I learned a lot about these things a few
weeks ago, since I live in Chicago, about 6 miles from Wrigley Field, in
a house full of Cubs fans.
S_A_V [24]3 years ago
6 0

Answer:

answer is B 19 newtons for plato users

Explanation:

got 100% on test, the other answer explains the method pretty well

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After the first few seconds of a race, a runner runs at the same speed until she approaches the finish line, at which point she
Rzqust [24]

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5 0
3 years ago
A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.
Neko [114]
In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

Substituting;
P_{2} = \frac{ P_{1}  T_{2} }{ T_{1} } = \frac{40*305}{289}  =42.21 psi
3 0
3 years ago
From the gravitational law calculate the weight W (gravitational force with respect to the earth) of a 89-kg man in a spacecraft
zhannawk [14.2K]

Answer:

W=\frac{773}{4.45}=173.76 l b f

Explanation:

W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}

The law of gravitation

G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)

Universal gravitational constant [S.I. units]

m_{e}=5.976\left(10^{24}\right) k g

Mass of Earth [S.I. units]

m=89 kg

Mass of a man in a spacecraft [S.I. units]

R=6371 \mathrm{~km}

Earth radius [km]

Distance between man and the earth's surface

h=261 \mathrm{~km} \quad[\mathrm{~km}]

ESULT W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}

W=\frac{773}{4.45}=173.76 l b f

4 0
2 years ago
An electric light is plugged into a 120-V outlet. If the current in the bulb is 0.50 A, how much electrical energy does the bulb
ioda

Answer:

= 54,000 Joules or 54 kJ

Explanation:

Electrical energy is given by the formula;

E = VIt; where V is the potential difference in volts, I is the current and t is the time in seconds.

Therefore;

Electrical energy = 120 V × 0.50 A × 15 ×60 seconds

                            = 54,000 Joules

Thus; the electrical energy is 54,000 joules or 54 kJ

7 0
3 years ago
Two waiters are trying to get through a single door of a kitchen. One pushes on one side of a door 0.567 m from the hinge with a
NISA [10]

Answer:

275.5 N

Explanation:

F_{1} = Force on one side of the door by first waiter = 257 N

F_{2} = Force on other side of the door by second waiter

r_{1} = distance of first force by first waiter from hinge = 0.567 m

r_{2} = distance of second force by second waiter from hinge = 0.529 m

Since the door does not move. hence the door is in equilibrium

Using equilibrium of torque by force applied by each waiter

r_{1} F_{1} = r_{2} F_{2} \\(0.567) (257) = (0.529) F_{2}\\F_{2} = 275.5 N

7 0
3 years ago
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