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lutik1710 [3]
3 years ago
14

Two students, standing on skateboards, are initially at rest, when they give each other a shove! After the shove, one student (7

7kg) moves to the left at 5.7m/s while the other student (59kg) moves right.
Physics
1 answer:
Elden [556K]3 years ago
6 0

Answer:

The other student (59kg) moves right at 7.44 m/s

Explanation:

Given;

mass of the first student, m₁ = 77kg

mass of the second student, m₂ = 59kg

initial velocity of the first student, u₁ = 0

initial velocity of the second student, u₂ = 0

final velocity of the first student, v₁ = 5.7 m/s left

final velocity of the second student, v₂ = ? right

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(77 x 0) + (59 x 0) = (-77 x 5.7) + (59 x  v₂)

0 = - 438.9 + 59v₂

59v₂ = 438.9

v₂ = 438.9 / 59

v₂ = 7.44 m/s to the right

Therefore, the other student (59kg) moves right at 7.44 m/s

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12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be
Nat2105 [25]

Answer:

81:256.

Explanation:

Let T denote the absolute temperature of this object.

Calculate the value of T before and after heating:

T(\text{before}) = 27 + 273 = 300\; \rm K.

T(\text{after}) = 127 + 273 = 400\; \rm K.

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to T^4.

Ratio between the absolute temperature of this object before and after heating:

\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}.

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

\displaystyle \left(\frac{T(\text{before})}{T(\text{after})}\right)^{4} = \left(\frac{3}{4}\right)^{4} = \frac{81}{256}.

4 0
3 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}

where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
on aircraft carriers, catapults are used to accelerate jet air craft to flight speeds in a short distance. One such catapult tak
sineoko [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

Acceleration = (70 / 2.5) m/s²

<em>Acceleration = 28 m/s²</em>  

That's about 2.9 G's .  Jet pilots can endure a lot more than that, but maybe the catapult or the hook on the airplane can't.  Let's look a little closer:

F = m A (Newton #2)

The force on the airplane = (18,000 kg) x (28 m/s²)

Force on the airplane = 504,000 Newtons

That's about 113,000 pounds !  Maybe the part of the airplane that the catapult pushes on can't handle any more force than that.  Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds.  Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes !  Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

8 0
3 years ago
To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance
mart [117]

Answer:

a. 120 W

b. 28.8 N

Explanation:

To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance. Suppose a cyclist is traveling at 15 km/h on level ground. Assume he is using 480 W of metabolic power.

a. Estimate the amount of power he uses for forward motion.

b. How much force must he exert to overcome the force of air resistance?

(a) He is 25% efficient, therefore the cyclist will be expending 25% of his power to drive the bicycle forward

Power = efficiency X metabolic power

= 0.25 X 480

= 120 W

(b)

power if force times the velocity

P = Fv

convert  15 km/h to m/s

v = 15 kmph = 4.166 m/s

F = P/v

= 120/4.166

= 28.8 N

definition of terms

power is the rate at which work is done

force is that which changes a body's state of rest or uniform motion in a straight line

velocity is the change in displacement per unit time.

3 0
3 years ago
Body composition refers to:
NISA [10]
Bone, fat and muscle
3 0
2 years ago
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