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2 years ago

When you catch a baseball with a glove (instead of your hand), the glove helps by

Physics

1 answer:

AveGali [126]2 years ago

4 0

Answer:D. Impulse

Explanation:

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A mass of 10 g of oxygen fill a weighted pistonâ€“cylinder device at 20 kPa and 110Â°C. The device is now cooled until the tempe

mezya [45]

Answer:

The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

$\Delta V=V_{2}-V_{1}$

$\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})$

Put the value into the formula

$\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)$

$\Delta V=14.297\ L$

$\Delta V=14.3\times10^{-3}\ m^3$

Hence, The change of the volume of the device during this cooling is $14.3\times10^{-3}\ m^3$

6 0

2 years ago

The process of sediments being compacted and cemented to form sedimentary rocks is called

crimeas [40]

Answer:

Lithification is the answer.

8 0

2 years ago

Helpppppppp please !!!!

agasfer [191]

Answer:

second one is correct that is right

7 0

3 years ago

The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x

Diano4ka-milaya [45]

Answer:

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Explanation:

Given that

At X=0 V=Vo

At X=X1 V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

$V^2=U^2-2aS$

$0=V_o^2-2a X_1$

$a=\dfrac{V_o^2}{2X_1}$

So the friction force on the box

Ff= m x a

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Where m is the mass of the box.

4 0

3 years ago

An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t

UkoKoshka [18]

Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

density of water, d = 1000 kg/m^3

Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = $1.013 \times 10^{5}+ 100 \times 1000 \times 9.8$

P' = 10.813 x 10^5 N/m^2

Use the Boyle's law

P V = P' V'

$1.013 \times 10^{5}\times V = 10.813 \times 10^{5}\times 2$

V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

5 0

2 years ago