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3 years ago

9

Developing

1 answer:

kow [346]3 years ago

6 0

m=1800 kg has the greatest momentum

<u>Explanation:</u>

Linear momentum is defined as the product of mass and velocity of a body. It is directly proportional to mass and velocity and is given by the expression

P=mv

where P denotes linear momentum,m denotes mass and v denotes velocity.

given that

velocity of both objects is v=5 m/s

Mass of the first object m= $1.2\times 10 kg$

Mass of the second object m=1800 kg

velocity of both objects is same. The second object has higher mass than first.

When mass increases, momentum increases with it.

Thus the second object with m=1800 kg has higher momentum than first.

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You need to repair a gate on the farm. The gate weighs 100 kg and pivots as indicated. A small diagonal bar supports the gate an

tekilochka [14]

Answer:

The force is $F = 3920 \ N$

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The weight of the gate is $G = 100\ kg$

The vertical component of F is $F_y = F\ sin \theta$

From the diagram , taking moment about the pivot we have

$W_g * 2 - F_y * 1 = 0$

Where $W_g$ is the weight of the gate evaluated as

$W_g = m_g * g = 100 * 9.8 = 980 \ N$

=> $980 * 2 - Fsin(30) * 1 = 0$

=> $F = \frac{1960}{sin(30)}$

=> $F = 3920 \ N$

7 0

3 years ago

The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he

schepotkina [342]

Answer:

The change in volume is $6.885\times 10^{- 5}\$

Solution:

As per the question:

Coefficient of linear expansion of Copper, $\alpha = 17\times 10^{- 6}\ K^{- 1}$

Initial Temperature, T = $0^{\circ}$ = 273 K

Final Temperature, T' = $100^{\circ}$ = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = $30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}$

$V' = V(1 + \gamma \Delta T)$

$\gamma = 3\alpha$

$V' = V(1 + 3\alpha \Delta T)$

where

V' = Final volume

$V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))$

$\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\$

7 0

3 years ago

What is in the universe

svp [43]

Everything is in the universe.
That's what the word means.

5 0

3 years ago

Read 2 more answers

What is the relationship between force in velocity selector in a bain bridge.

kifflom [539]

Answer:

In a velocity selector, there are two forces namely;

» Electric field Intensity

» Magnetic field density

<u>Relationship</u><u>:</u>

$E _{e}= B$

E is the electric field intensity

B is the magnetic flux density

4 0

3 years ago

*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co

sp2606 [1]

Answer: The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

$I = \sqrt{\frac{P}{R} }$

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

$I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }$

Then, simplify the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} }$

Rationalize the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}$

Simplify the numerator by finding its factors:

$I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}$

The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

8 0

3 years ago