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sveticcg [70]
3 years ago
7

A carver begins work on the following block of granite that weighs 2700 g. The dimensions of the block are 20 cm, 5 cm and 10 cm

. What is the density of the granite in g/cm³?
Physics
1 answer:
CaHeK987 [17]3 years ago
6 0

Answer:

\text { "Density" of the given granite block is } 2.7 \mathrm{g} / \mathrm{cm}^{3} \text { . }

Explanation:

Density of an object is defined as the mass occupied per unit volume. “Density” is represented by “ρ”. Mathematically, \rho=\frac{m}{v} where, “m" mass of an object and "v" is volume of an object.

Given that,

Dimensions of the block are 20 cm, 5 cm and 10 cm.

\text { Volume of the block is } l \times b \times h

\text { Volume of the block is } 20 \times 5 \times 10

\text { Volume of the block is } 1000 \mathrm{cm}^{3} \text { . }

Mass of the block is 2700 g.

Substitute the given values in \rho=\frac{m}{v} to find density of a block.

\rho=\frac{2700}{1000}

\rho=2.7 \mathrm{g} / \mathrm{cm}^{3}

\text { "Density" of the given granite block is } 2.7 \mathrm{g} / \mathrm{cm}^{3}

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The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
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Part a

Answer: NO

We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.

Using the second equation of motion:

s=ut+0.5at^2

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.

It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.

Part b

Answer: 29.6 m/s

The maximum distance that car can travel is s=44.0 m

The acceleration is same, a=-10.0 m/s^2

The final velocity, v=0

Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.





7 0
3 years ago
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