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Stella [2.4K]
3 years ago
6

Calculate the mass of a 0.9 m^3 block of a material having a density of 12500 kg/m^3.

Physics
1 answer:
timama [110]3 years ago
8 0

Answer: The mass of the object will be 11250 kg.

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given :

Density of the object= 12500kg/m^3

Mass of object = ?

Volume of the object = 0.9m^3

Putting in the values we get:

12500kg/m^3=\frac{mass}{0.9m^3}

12500kg/m^3=\frac{mass}{0.9m^3}

mass=11250kg

Thus the mass of the object is 11250 kg.

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A billiard ball traveling at 3.4 m/s has an elastic head-on collision with a billiard ball of equal mass that is initially at re
Tpy6a [65]

Answer:

Explanation:

Given

Initial velocity of first billiard ball u_1=3.4\ m/s

Initial velocity of second billiard ball u_2=0\ m/s

After collision first ball comes to rest

suppose m is the mass of both the balls

Conserving momentum to get the speed of second ball after collision

Initial momentum P_i=mu_1+mu_2=m\cdot 3.4+0

Final momentum P_f=mv_1+mv_2

where v_1 and v_2 are the speed of first and second ball respectively

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m\cdot 3.4+0=0+m\cdot v_2

v_2=3.4\ m/s

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3 years ago
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A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?
lapo4ka [179]
<h2>Answer: 12.24m/s</h2>

According to <u>kinematics</u> this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

V_{f}^{2}-V_{o}^{2}=2.a.d   (1)

Where:

V_{f} is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.

V_{o} is the Initial Velocity, the value we want to find

a is the constant acceleration of the car (the negative sign means the car is decelerating)

d is the distance traveled by the car

Now, let's substitute the known values in equation (1) and find V_{o}:

0-V_{o}^{2}=2(-5m/{s}^{2})(15m)    (2)

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Multiplying by -1 on both sides of the equation:

V_{o}^{2}=150{m}^{2}/{s}^{2}    (4)

V_{o}=\sqrt{150{m}^{2}/{s}^{2}}    (5)

Finally:

V_{o}=12.24m/s >>>This is the Initial velocity of the car

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A projectile, fired with unknown initial velocity, lands 20sec later on side of hill, 3000m away horizontally and 450m verticall
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Explanation:

Given:

t = 20 seconds

x = 3000 m

y = 450 m

a) To find the vertical component of the initial velocity v_{0y}, we can use the equation

y = v_{0y}t - \frac{1}{2}gt^2

Solving for v_{0y},

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\:\:\:\:\:\:\:=\dfrac{(450\:\text{m}) + \frac{1}{2}(9.8\:\text{m/s}^2)(20\:\text{s})^2}{(20\:\text{s})}

\:\:\:\:\:\:\:=120.5\:\text{m/s}

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x = v_{0x}t \Rightarrow v_{0x} = \dfrac{x}{t} = \dfrac{3000\:\text{m}}{20\:\text{s}}

or

v_{0x} = 150\:\text{m/s}

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