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Vikentia [17]
4 years ago
11

A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa

rt A What will be their mutual speed immediately after the collision? Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
DIA [1.3K]4 years ago
3 0

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

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In the absence of an electric field, a radioactive beam strikes a fluorescent screen at a single point. When an electric field i
leonid [27]

Answer: Beta, alpha and gamma ray

Explanation: The component being deflected to the positive side is the beta radiation because it is negatively charged and thus is attracted by the positive terminal of the Electric Field.

The component being deflected to the negative side is the alpha radiation, it's is positively charged and thus being attracted by the negative part of the Electric Field.

The component that went straight down without deflection is the gamma radiation. It is neutral and possess no charge and thus is not deflected.

4 0
3 years ago
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Why is it possible to throw a 0.145 kg baseball much further than a 7 kg bowling ball
Lilit [14]
Simply because the baseball is much lighter than the bowling ball. Making it easier for the average adult to throw the baseball farther than the bowling ball
5 0
3 years ago
A copper sphere was moving at 25 m/s when it hit another object. this caused all of the ke to be converted into thermal energy f
muminat

ΔT= 81°C  was the increase in temperature.

<h3>steps</h3>

As evidenced by energy conservation

To increase its temperature, all of its kinetic energy will be converted to thermal energy.

\frac{1}{2}mv ^{2} = msΔT

then divide both sides by the object's mass.

\frac{1}{2} v^{2} =sΔT

therefore, a temperature shift is described as

ΔT= V^{2} /2s

ΔT= 25^{2}/ 2 x 387

ΔT= 81°C

ΔT= 81°C  was the increase in temperature.

learn more about increase in temperature here

<u>brainly.com/question/14319779</u>

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3 0
1 year ago
a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
3 years ago
A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di
KATRIN_1 [288]

Answer:

C. 1400

Explanation:

The force exerted = f1 = 250

Diameter d1 = 0.02

r1 = d1/2 = 0.01

Diameter d2 = 0.15

r2 = d2/2 = 0.075

The mass of jack to lift

F1/A1 = f2/A2

250/r1² = f2/r2²

250/0.01² = f2/0.075²

When we cross multiply, we will have:

250x0.005625 = 0.0001f2

1.40625 = 0.0001f2

F2 = 1.40625/0.0001

F2 = 14062.5N

Force f = mg

g = 9.81

m = 14062.5/9.81

Mass = 1433Kg

Therefore option c is correct. A mass of 1400kg can be lifted by the jack

7 0
3 years ago
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