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3 years ago

A cannon fires a shell toward a target with a momentum of 6750 kg m/s. This shell moves with a velocity of 150 m/s. Calculate

Physics

1 answer:

Juliette [100K]3 years ago

3 0

Answer:

Explanation:

The mass of the shell can be found by using the formula

$m = \frac{p}{v} \\$

p is the momentum

v is the velocity

From the question we have

$m = \frac{6750}{150} = \frac{675}{15} \\$

We have the final answer as

Hope this helps you

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Light is reflected from a crystal of table salt with an index of refraction of 1.544. An analyser is placed to intercept the ref

Vitek1552 [10]

Answer:

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Explanation:

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5 0

3 years ago

The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth

FinnZ [79.3K]

Answer:

the pressure at the depth is 1.08 × $10^{8}$ Pa

Explanation:

The pressure at the depth is given by,

P = h $\rho$ g

Where, P = pressure at the depth

h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter

$\rho$ = density of water = 1000 $\frac{kg}{m^{3} }$

g = acceleration due to gravity ≈ 9.8 $\frac{m}{s^{2} }$

P = 11033.15 × 9.8 × 1000

P = 1.08 × $10^{8}$ Pa

Thus, the pressure at the depth is 1.08 × $10^{8}$ Pa

4 0

3 years ago

A fire helicopter carries a 580-kg bucket of water at the end of a 20.0-m long cable. Flying back from a fire at a constant spee

inn [45]

Answer:

F = 41,954 N

Explanation:

given,

mass of bucket = 580 Kg

length of the cable = 20 m

velocity = 40 m/s

angle made = 38.0°

T cos 38° = m g..............(1)

T sin 38^0 = \dfrac{mv^2}{l} + F......(2)

dividing equation (2) by (1)

$tan 38^0 = \dfrac{\dfrac{mv^2}{l} + F}{mg}$

$tan 38^0 = \dfrac{\dfrac{580\times 40^2}{20} + F}{580 \times 9.81}$

$4445.36 = \dfrac{580\times 40^2}{20} + F$

F = -46400 + 4445.36

F = -41,954 N

hence, the force is acting in the opposite direction as assumed.

F = 41,954 N

4 0

3 years ago

The 1.0-kg collar slides freely on the fixed circular rod. Calculate the velocity v of the collar as it hits the stop at B if it

soldi70 [24.7K]

Answer:

6.21 m/s

Explanation:

Using work energy equation then

$U_{1-2}=T_B- T_A\\58d-mgh=0.5m(v_b^{2}-v_a^{2})$

where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity

Substituting 1 Kg for m, 0.4m for h, $v_a$ as 0, 9.81 for g then

$58(\sqrt{0.4^{2}+0.3^{2}}-0.1)-(1\times 9.81\times 0.4)=0.5\times 1\times (v_b^{2}-v_a^{2})\\19.276=0.5\times 1v_b^{2}\\v_b=6.209025688 m/s\approx 6.21 m/s$

7 0

3 years ago

The electric field strength E₀ is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a un

Troyanec [42]

Answer:

Explanation:

E=(σ/ε0)

As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."

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3 years ago