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2 years ago

Use the image to determine the volume of the rock.

Physics

1 answer:

maria [59]2 years ago

6 0

Answer:

12 ml

Explanation:

The initial volume in the cylinder is 20 ml

adding the rock adds volume to the cylinder

the new volume is 32 ml .....the increase in volume is the volume of the rock : 32 - 20 = 12 ml volume of rock

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Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 2.20×106 N , one at an angle 19.0 ∘ west of north,

valina [46]

Answer:

$=2.99\times10^9J$

Explanation:

According to the question

net force F = 2.20×10^6 N

displacement $S = 0.72\times10^3m$

from figure , the horizontal forces are same in magnitude and opposite direction.

so , neglect these two forces.

we can take only vertical components of the force.

total force F' = F cos 19° + F cos 19°

= 2×F×cos 19° ................. (1

therefore , total work is

W = F'S

= (2F cos19)×S

$= (2)(2.20\times10^6 N)cos19° (0.720\times10^3 m)$

$=2.99\times10^9J$

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3 years ago

Can you make sunglasses from new 3D-glasses???

katen-ka-za [31]

Yes you can make sunglasses from 3d glasses

5 0

3 years ago

Read 2 more answers

A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if

77julia77 [94]

C because when the part gets out of the probe it would no longer stay contacted

6 0

3 years ago

A particle moving with initial velocity of 5m/s is subjected to a uniform acceleration of -2.5 m/s2.find the displacement for ne

LUCKY_DIMON [66]

1/2vt^2-1/2at^2
1/2*5*16-1/2*2,5*16=20

6 0

3 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

3 years ago