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3 years ago

The volume of the gas at a temperature of 50 Kelvin on this scale is closest to

Physics

1 answer:

8_murik_8 [283]3 years ago

4 0

<span>500 cubic centimeters</span>

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fired, the ball moves 15.9 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.

sergeinik [125]

Answer:

Explanation:

Where is the remaining part of the question? This is a very easy one

6 0

2 years ago

Copper has a modulus of elasticity of 110 GPa. A specimen of copper having a rectangular cross section 15.2 mm × 19.1 mm is pull

bearhunter [10]

Answer:

$strain = 1.4 \times 10^{-3}$

Explanation:

As we know by the formula of elasticity that

$E = \frac{stress}{strain}$

now we have

$E = 110 GPA$

$E = 110 \times 10^9 Pa$

Area = 15.2 mm x 19.1 mm

$A = 290.3 \times 10^{-6}$

now we also know that force is given as

$F = 44500 N$

here we have

stress = Force / Area

$stress = \frac{44500}{290.3 \times 10^{-6}}$

$stress = 1.53 \times 10^8 N/m^2$

now from above formula we have

$strain = \frac{stress}{E}$

$strain = \frac{1.53 \times 10^8}{110 \times 10^9}$

$strain = 1.4 \times 10^{-3}$

7 0

3 years ago

A weightless spring is stretched 10 cm by a suspended 1-kg block. If two such springs are used to suspend the block, one spring

leva [86]

Answer:

total stretch of the double-length spring will be 20 cm

Explanation:

given data

length x1 = 10 cm

mass = 1 kg

mass = double = 2 kg

to find out

the total stretch of the double-length spring will be

solution

we can say here spring constant is

k = mg ............1

k is spring constant and m is mass and g is acceleration due to gravity

so for in 1st case and 2nd case with 1 kg mass and 2 kg mass

kx1 = mg .........................2

and

kx2 = 2mg ........................3

x is length

so from equation 2 and 3

$\frac{kx1}{kx2}= \frac{1mg}{2mg}$

$\frac{x1}{x2} = \frac{1}{2}$

$\frac{10}{x2} = \frac{1}{2}$

x2 = 20

so total stretch of the double-length spring will be 20 cm

7 0

3 years ago

A car drives at a constant speed of 21 m/s around a circle of radius 100 m.

Arte-miy333 [17]

Answer:

Option D. 4.4 m/s²

Explanation:

The following data were obtained from the question:

Velocity (v) = 21 m/s

Radius (r) = 100 m

Centripetal acceleration (a) =.?

The centripetal acceleration of the car can be obtained as follow:

Centripetal acceleration (a) = Velocity square (v²) / radius (r)

a = v²/r

a = 21²/100

a = 441/100

a = 4.41 ≈ 4.4 m/s²

Therefore, the centripetal acceleration of the car is 4.4 m/s².

8 0

3 years ago

A diatomic molecule of Nitrogen (N) would be held by covalent bonds.

zzz [600]

Yes that would be true

3 0

2 years ago

Read 2 more answers