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2 years ago

Sean pushes a box with a force of 100 N. Christian pushes an identical box with a force of

Physics

1 answer:

kap26 [50]2 years ago

5 0

Answer:

Christian

Explanation:

Given that

Force applied by Sean ,F₁=100 N

Force applied by Christian ,F₂=200 N

Lets take mass of the box = m kg

As we know that from second law of Newton's

F= m a

F=Force,m=mass ,a=acceleration

$F_1=ma_1$

$100=m\times a_1$

$a_1=\dfrac{100}{m}\ m/s^2$

$F_2=ma_2$

$200=m\times a_2$

$a_2=\dfrac{200}{m}\ m/s^2$

From the above we can say that acceleration a₂ is greater than a₁ is for the same mass of the box.

Therefore Christian will acceleration more for the box.

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The universe was 5 percent its current size when light left objects observed now at redshift of ______________

Nuetrik [128]

The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.

The Universe was 5 percent its current size when light left objects now at redshift of 19.

Reasons:

The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

$\displaystyle \frac{a}{a_0} =\mathbf{ \frac{1}{1 + z}}$

Where;

a₀ = The current size of the Universe

a = The size of the early Universe = 5% of a

Therefore;

$\displaystyle \frac{a}{a_0} =5\% = 0.05= \frac{1}{1 + z}$

$\displaystyle 0.05 = \frac{1}{1 + z}$

0.05 + 0.05·z = 1

$\displaystyle z = \mathbf{ \frac{1 - 0.05}{0.05} } = 19$

The redshift is of the observed light is, z = 19

Learn more here:

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4 0

2 years ago

How much work is done when a 100 N force moves a block 59 m?

xxTIMURxx [149]

Answer:

5900J

Explanation:

Work=Forse*Distance

work = J, Jewls

100*59=5900

Hop this helps and can u think about brainlist

i put a picture on how to find these answers, if u got any more questions im here

3 0

3 years ago

NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

2 years ago

On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The

Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

$a = g (\sin \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2$

Velocity at the bottom

$v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s$

after travelling 4m , its velocity becomes 0

$a=\frac{v^2-u^2}{2s}$

$a=\frac{0-u^2}{2s}$

$a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2$

Coefficient of kinetic friction

μ = F/N

$=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31$

Therefore, the Coefficient of kinetic friction is 0.31

8 0

2 years ago

Read 2 more answers

if you run off the pavement, you should: turn the steering wheel quickly toward the road steer straight and slow down before att

omeli [17]

pavement is defined as the surface of Road or sidewalk.

for example, the surface of Expressway.

There are two types of pavement.

rigid pavement which consists of one layer.

flexible pavement which consist of multiple layers.

While driving on roads of rural areas, if our right wheel moves off the pavement, we should always hold the steering wheel firmly and then take our foot off the pedal, then apply brake lightly until we are moving at a low speed.

if you run off the pavement, you should: turn the steering wheel quickly toward the road steer straight and slow down before attempting to return to the pavement steer straight ahead and speed up apply the brakes hard

To know more about pavement:

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5 0

1 year ago