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Naily [24]
3 years ago
15

A ball of mass 0.7 kg flies through the air at low speed, so that air resistance is negligible. (a) What is the net force acting

on the ball while it is in motionWhich components of the ball's momentum will be changed by this force? What happens to the x component of the ball's momentum during its flight? What happens to the y component of the ball's momentum during its flight? It decreases. What happens to the z component of the ball's momentum during its flight?
Physics
1 answer:
Fed [463]3 years ago
5 0

Answer:

A) Gravitational Force, F=-6.86N.

B) The y component of momentum.

C) The x component of momentum will remain the same.

D)The y component of momentum decreases.

E)The z component of momentum will remain constant.

Explanation:

First, let the x-axis be positive in the same direction as the horizontal speed of the ball, let the y-axis be positive in the opposite direction of gravity.

A) since there is no air resistance, the only force acting on the ball is the gravitational force. This force is downwards so it is negative. The net force is

F=-9.8*0.7=-6.86N.

B)Because the net force (and the only one) is on the y-axis, only the vertical component of the momentum will be changed due to the force.

C)Since there is no resistance of air, the ball behaves as in projectile motion problems, this means that the x component of the velocity remains constant, also does the mass. Remembering that momentum is velocity times mass, it can be easily seen that x component of momentum remains constant too.

D)The y component of momentum decreases, this is because gravity decreases the y component of the velocity.

E)Since there is no z component of the force there is no change in the z component of the momentum.

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21 kg brick v=6.1 m/s sticks to a 2.9 kg cart v=.26 m/s
vovikov84 [41]

Answer:

v = 6.05 m/s

Explanation:

The mass of the brick, M = 21 kg

The velocity of the brick, U = 6.1 m/s

The mass of the cart, m = 2.9 kg

The velocity of the cart, u = 0.26 m/s

The combined mass of the brick and the cart, v = ?

According to the law of conservation of linear momentum,

         Combined final momentum = sum of the initial momentum

                         (M + m)v = MU + mu

                                      v = MU + mu / M + m

                                         = 21 x 6.1 + 2.1 x 0.26 / (21 + 0.26)

                                         = 6.05 m/s

Hence, the combined velocity of the cart and the brick is, v = 6.05 m/s

5 0
3 years ago
Suppose that an object undergoes simple harmonic motion, and its displacement has an amplitude A = 15.0 cm and a frequency f = 1
cestrela7 [59]

Answer:

Maximum speed ( v ) = 10.4 m/s (Approx)

Explanation:

Given:

Amplitude A = 15.0 cm = 0.15 m

Frequency f = 11.0 cycles/s (Hz)

Find:

Maximum speed ( v )

Computation:

Angular frequency = 2πf

Angular frequency = 2π(11)

Angular frequency = 69.14

Maximum speed ( v ) = WA

Maximum speed ( v ) = 69.14 x 0.15

Maximum speed ( v ) = 10.371

Maximum speed ( v ) = 10.4 m/s (Approx)

5 0
3 years ago
One component of a metal sculpture consists of a solid cube with an edge of length 24.5 cm . The alloy used to make the cube has
boyakko [2]

Answer:

118kg

Explanation:

answered

Given

density of the cube= 8050 kg/m3 .

length of the sizes of the cube=24.5 cm

We can convert the length to cm for unit consistency.

It's length =24.5 cm =0.245m

✓ the length of sizes of the cube is the same, then the volume can be calculated as

Volume= L^3

= (0.245m)^3

=0.01470625 m^3

✓ but we know that

Density = mass/ volume

Then,

Mass= (Volume × density)

= (0.01470625)(8050)

= 118 kg

Hence, the mass of the cube is 118 kg

8 0
3 years ago
The water that powers the generators enters and leaves the system at a low speed (thus we can neglect its change of kinetic ener
madam [21]

Answer:

Explanation:

Let density of water be ρ .

During flow , volume of water flowing per second is constant

loss of P. E per unit volume  = ρ gh , 83.5 % is lost

Gain of K E per unit volume = 1/2  ρ v²

83.5 % of mgh =   ρ 1/2  ρ v²

1/2  ρ v² =  .835 x 9.8

v² =  2 x .835 x 9.8

= 16.366

v = 4.04 m /s

4 0
4 years ago
Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the
Vinil7 [7]

Answer:

a) The exit temperature is 430 K

b) The inlet and exit areas are 0.0096 m² and 0.051 m²

Explanation:

a) Given:

T₁ = 127°C = 400 K

At 400 K, h₁ = 400.98 kJ/kg (ideal gas properties table)

The energy equation is:

q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2}    }{2} +delta-p

For a diffuser, w = Δp = 0

The diffuser is adiabatic, q = 0

Replacing:

0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2}    }{2} +0

Where

V₁ = 250 m/s

V₂ = 40 m/s

Replacing:

0-0=h_{2} -400x10^{3}  +\frac{40^{2}-250^{2}      }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg

Using tables, at 431.43 kJ/kg the temperature is 430 K

b) The inlet area is:

m=\frac{P_{1} }{RT_{1} } A_{1} V_{1} \\\frac{6000}{3600} =\frac{80}{0.287*400} A_{1} *250\\A_{1} =0.0096m^{2}

The exit area is:

m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}

6 0
3 years ago
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