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Drupady [299]
2 years ago
12

Volume of an block is 5 cm3. If the density of the block is 250 g/cm3, what is the mass of the block ?​

Physics
1 answer:
wlad13 [49]2 years ago
7 0

Answer:

The mass of the block is 1250g.

Explanation:

Given that the formula for density is ρ = mass/volume. Then you have to substitute the values into the formula :

ρ =  \frac{mass}{volume}

Let density = 250,

Let volume = 5,

250 =  \frac{m}{5}

m = 250 \times 5

m = 1250g

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Explanation:

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A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 an
Tom [10]

Answer:

0.69s

Explanation:

10 cm = 0.1 m

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

\omega = \alpha t = 2.1 t

And so the radial acceleration is

a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2

The tangential acceleration is always the same since angular acceleration is constant:

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For these 2 quantities to be the same

a_r = a_t

0.441 t^2 = 0.21

t^2 = 0.21/0.441 = 0.4762

t = \sqrt{0.4762} = 0.69 s

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3 years ago
IF a car trip 400 km in four hours what is the rate of speed that a car is traveling?
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Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0
2 years ago
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