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gregori [183]
4 years ago
5

An arrow of mass 415 g is shot at a target with a speed of 68.5m/s. The target, which has a mass of 3.3 kg, is moving toward the

arrow with a speed of 1.1m/s. The surface on which the target is sliding is friction-free. If the arrow embeds itself in the target, what will be the velocity of the target/arrow mass after the collision? Take the arrow's initial velocity direction as positive
Physics
1 answer:
Murrr4er [49]4 years ago
4 0

The velocity of the target and arrow after collision is 6.67m/s

<u>Explanation:</u>

Given:

Mass of arrow, mₐ = 415g

Speed of arrow, vₐ = 68.5m/s

Mass of the target, mₓ = 3.3kg = 3300g

speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction

Velocity of the target and arrow after collision, vₙ = ?

Applying the conservation of momentum,

mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ

415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ

28427.5 - 3630 = 3715 X vₙ

24797.5 = 3715 X vₙ

vₙ = 6.67m/s

Therefore, the velocity of the target and arrow after collision is 6.67m/s

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A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which
julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = 11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = 390\ W/m.^{\circ}C

Conduction of heat from the rod per second is given by:

q = \frac{KA\Delta T}{L}

where

\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C = temperature difference between the two ends of the rod.

Thus

q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s

Now,

To calculate the mass, M of the ice melted per sec:

M = \frac{q}{L_{w}}

where

L_{w} = Latent heat of fusion of water = 333 kJ/kg

M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g

5 0
3 years ago
Question 6. B) and c)
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The formula for both is v(t) = v0 + a*t

b) v(8) = 0 + 6m/s^2 *8s = 48 m/s

now we know the beginning (2) and end speed (14), but not the time:

c) 14 = 2 + 1.5*t => t = (14-2)/1.5 = 8 seconds
4 0
4 years ago
At Hoover Dam, the distance the water effectively falls before encountering the electric generators depends on the water levels
pashok25 [27]

Answer:

Explanation:

recall that power is energy carried out or work done per time

P=W/t

P=2*10^6*35

t=6*60=420S

W=Energy

E=2*10^6*35*360S

E=25200000000

Energy stored by water from rest is called potential energy. Since the water is falling from a height , we calculate potential energy as thus

E=M*g*h

Assume that the water intakes are effectively 175 m above the electric generators. How much water must pass through the generators to power 2 million 35-W Las Vegas light bulbs for 6.0 minutes?

M=mass of water

g=acceleration due to gravity 9.81m/s^2

h=height ,175m

25200000000=M*9.81*175

M=\frac{25200000000}{175*9.81}

M=1716.75kg

3 0
3 years ago
Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

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6 0
4 years ago
An electric motor rotating a workshop grinding wheel at 1.00 × 10² rev/min is switched off. Assume the wheel has a constant nega
ololo11 [35]

An electric engine turning a workshop sanding rotation at 1.00 × 10² rev/min is switched off. Take the wheel includes a regular negative angular acceleration of volume 2.00 rad/s². 5.25 moments long it takes the grinding rotation to control.

<h3>What is negative angular acceleration?</h3>
  • A particle that has a negative angular velocity rotates counterclockwise.
  • Negative angular acceleration () is a "push" that is hence counterclockwise.
  • The body will speed up or slow down depending on whether and have the same sign (and eventually go in reverse).
  • For instance, when an object rotating counterclockwise slows down, acceleration would be negative.
  • If a rotating body's angular speed is seen to grow in a clockwise direction and decrease in a counterclockwise direction, it is given a negative sign.
  • It is known that a change in the linear acceleration correlates to a change in the linear velocity.

Let t be the time taken to stop.

ω = 0 rad/s

Use the first equation of motion for rotational motion

ω = ωo + α t

0 = 10.5 - 2 x t

t = 5.25 second

To learn more about angular acceleration, refer to:

brainly.com/question/21278452

#SPJ4

7 0
2 years ago
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