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2 years ago

A sled starts from rest,

Physics

1 answer:

Scorpion4ik [409]2 years ago

6 0

Answer:

The acceleration of the sled is $4.6\ m/s^2$ .

Explanation:

It is given that,

Initial speed of sled is 0 because it was at rest.

It is placed at an angle of 28° on a frictionless hill.

We need to find the acceleration of the sled. It is placed at an incline. It means that the acceleration of the sled is given by :

$a=g\sin\theta\\\\a=9.8\times \sin(28)\\\\a=4.6\ m/s^2$

So, the acceleration of the sled is $4.6\ m/s^2$ .

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A substance is soluble in water. It is added into the water and dissolves. More solute continues being added until it reaches a

Sophie [7]

Answer:

option D

Explanation:

also to add.... there exists a solubility equilibrium between solid solute and the solution.

4 0

3 years ago

I have been trying to do number 2 and I just don’t know what the answer is if you can help me thank you so mush

skelet666 [1.2K]

Answer:

Strong Positive Linear Line Correlation

Explanation:

According to the diagram, there is a strong positive linear line correlation between velocity and distance of galaxies. Hubble showed that galaxies are receding away from people with a velocity that is proportional to their distance from people more distant galaxies recede faster than nearby galaxies.

About the diagram: Velocity is the distance relation among extra-galactic nebulae. Radial velocities is corrected for solar motion, are plotted against distances estimated from involved stars and mean luminosities of nebulae in a cluster. The black discs and full line represent the solution for solar motion by using the nebulae individually; the circles and broken line represent the solution combining the nebulae into groups; the cross represents the mean velocity corresponding to the mean distance of 22 nebulae whose distances could not be estimated individually.

4 0

2 years ago

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

Olegator [25]

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

3 0

3 years ago

A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 x 106 meters, find the mass of

melisa1 [442]

Answer:

7.36 × 10^22 kg

Explanation:

Mass of the man = 90kg

Weight on the moon = 146N

radius of the moon =1.74×10^6

Weight =mg

g= weight/mass

g= 146/90 = 1.62m/s^2

From the law of gravitational force

g = GM/r^2

Where G = 6.67 ×10^-11

M = gr^2/G

M= 1.62 × (1.74×10^6)^2/6.67×10^-11

= 4.904×10^12/6.67×10^-11

=0.735×10^23

M= 7.35×10^22kg. (approximately) with option c

3 0

2 years ago

Read 2 more answers

a force is applied to a block causing it to accelerate along a horizontal, frictionless surfaces. the energy gained by the block

GaryK [48]

Energy given to the block

3 0

2 years ago