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3 years ago

what is the acceleration of a softball if it has a mass of 0.50kg and hits the catcher’s glove with a force of 25 N

Physics

1 answer:

Kitty [74]3 years ago

3 0

Answer:

mass=0.50kg

force=25N

acceleration =?

Now,

force=m×a

25=0.50×a

25÷0.50=a

50=a

acceleration =50m/s^2 answer!!!!

hope this may help you!!!!

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At a particular instant, a proton at the origin has velocity < 5e4, -2e4, 0> m/s. You need to calculate the magnetic field

vesna_86 [32]

Answer:

$9.7\times 10^{-5} T$

Explanation:

Velocity = $5\times 10^4i-2\times 10^4j$

r= $0.03i+0.05j$

r= $\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058$

v= $\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}$

We know that

$B=\frac{mv}{qr}$

Where q= $1.6\times 10^{-19} C$

Mass of proton= $1.67\times 10^{-27} kg$

Using the formula

$B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}$

$B=9.7\times 10^{-5} T$

3 0

3 years ago

If it takes 600 N to move a box 5 meters, how much work is done on the box?

Anika [276]

Answer:

Option C is the correct option.

Explanation:

Given,

Force = 600 N

Distance = 5 meters

Work = ?

Now,

Work = Force $\times$ distance

$= 600 \times 5$

Calculate the product

$= 3000 \:$ Joule

Hope this helps...

Good luck on your assignment..

3 0

3 years ago

Read 2 more answers

I need help with 1-10

e-lub [12.9K]

Answer:

1) C. Energy

2) A. Joule

3) D. Joule

4) B. Potential

5) A. Greater

6) C. Largest

7) A. Speed and mass

8) A. Kinetic

9) A. Kinetic

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3 0

2 years ago

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

$P_{1}V_{1} = P_{2}V_{2}$

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

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3 years ago

A sailor pulls a crate across the deck of a ship with a rope, exerting a horizontal force of 150. N. The crate, which has a mass

nata0808 [166]

Answer:

Explanation:

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