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Vika [28.1K]
3 years ago
7

A box accidentally drops from a truck traveling at a speed of 20.0 m/s and slides along the ground for a distance of 30.0 m Calc

ulate a) The acceleration of the box (assuming it is constant while sliding)
Physics
1 answer:
Hoochie [10]3 years ago
6 0

Answer:

a= - 6.667 m/s²

Explanation:

Given that

The initial speed of the box ,u= 20 m/s

The final speed of the box ,v=  0 m/s

The distance cover by box ,s= 30 m

Lets take the acceleration of the box = a

We know that

v²= u ² + 2 a s

Now by putting the values in the above equation we get

0²=20² + 2 a x 30

a=- \dfrac{20^2}{2\times 30} \ m/s^2

a= - 6.667 m/s²

Negative sign indicates that velocity and acceleration are in opposite direction.

Therefore the acceleration of the box will be  - 6.667 m/s² .

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In a particular run of the “Archimedes principle” experiment a particular unknown substance is found to have a “dry mass” of 4.5
iris [78.8K]

Answer:

D = 18000 kg/m3

V = 2.5*10{-7}m3

Explanation:

From the Archimedes principle,

Weight of fluid displaced = W_{air} - W_{water}

W_{air} = 4.5 gm

W_{water} = 4.25 gm

W = [4.5 - 4.25]*9.81*10^{-3}

W = 2.4525*10{-3} N

\frac{density\ of\ object}{density\ of\ fluid} = \frac{weight\ in\ air}{weight\ of\ displaced\ fluid}

Density\ of\ object = \frac{D_{water}*Weight\ in\ air}{weight\ of\ displaced\ water}

D = \frac{1000*4.5*10^{-3}*9.8}{2.4525*10^{-3}}N

D = 18000 kg/m3

b) object Volume can be obtained  as ,

V = \frac{m}{D} = \frac{4.5*10^{-3}}{18000}

V = 2.5*10{-7}m3

7 0
3 years ago
From Kepler's third law, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to:
bazaltina [42]

Answer:

The correct option is (B).

Explanation:

The Kepler's third law of motion gives the relationship between the orbital time period and the distance from the semi major axis such that,

T^2\propto a^3\\\\T^2=ka^3

It is mentioned that, an asteroid with an orbital period of 8 years. So,

(8)^2=ka^3\\\\64=ka^3\\\\a=(64)^{\dfrac{1}{3}}\\\\a=4\ AU

So, an asteroid with an orbital period of 8 years lies at an average distance from the Sun equal to 4 astronomical units.

7 0
3 years ago
Look up the specs on a c6 rocket engine. How many C6 engines would it take to launch Mr. Blazey (90kg)
taurus [48]

Answer:

The Estes C6-0 engine is a booster stage engine designed for model rocket flight and has to be used with a standard engine. This engine is for flights in rockets weighing less than 4 ounces, including the engines. Each package includes 3 engines, 4 starters and 4 plugs.

3 0
2 years ago
Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling witho
8090 [49]

Answer:

Hoop will reach the maximum height

Explanation:

let the mass and radius of solid ball, solid disk and hoop be m and r  (all have same radius and mass)

They all  are rolled with similar initial speed v

by the law of conservation of energy we can write

K_{trans}+K_{rot}= P

for solid ball

[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{ball}\omega^2= mgh_{ball}

putting I_{ball}=\frac{2}{5}mr^2 and \omega=\frac{v}{r} in the above equation and solving we get

h_{ball}= 0.071v^2

now for solid disk

[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{disk}\omega^2= mgh_{disk}

putting I_{ball}=\frac{1}{2}mr^2 and \omega=\frac{v}{r} in the above equation and solving we get

h_{disk}= 0.076v^2

for hoop

[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{hoop}\omega^2= mgh_{hoop}

putting I_{hoop}=mr^2 and \omega=\frac{v}{r} in the above equation and solving we get

h_{hook}= 0.10v^2

clearly from the above calculation we can say that the Hoop will reach the maximum height

5 0
3 years ago
What is the law of reflection?
tia_tia [17]
You can observe this law in practice if <span>a ray of light reflects off of a flat mirror. 
</span>
Law of reflection states that both direction of both incoming and outgoing rays of light make the same angle with respect to surface normal.
4 0
3 years ago
Read 2 more answers
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