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DanielleElmas [232]

3 years ago

15

if a bullet fired beneath the water, straight up, breaks through the surface at a speed of 25m/s, how high will the bullet be af

ter 2.8 seconds? what is the bullet's velocity at this time? (accelaration of 9.8m/s)

1 answer:

Doss [256]3 years ago

8 0

Answer:

32 m and -2.4 m/s

Explanation:

Given:

v₀ = 25 m/s

t = 2.8 s

a = -9.8 m/s²

Find: Δy, v

Δy = v₀ t + ½ at²

Δy = (25 m/s) (2.8 s) + ½ (-9.8 m/s²) (2.8 s)²

Δy = 31.6 m

v = at + v₀

v = (-9.8 m/s²) (2.8 s) + 25 m/s

v = -2.44 m/s

Rounded to two significant figures, the bullet reaches a height of 32 m and a velocity of -2.4 m/s.

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A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

ArbitrLikvidat [17]

Answer:

The mass of the solid cylinder is $m = 1612.5 \ kg$

Explanation:

From the question we are told that

The radius of the grinding wheel is $R = 0.330 \ m$

The tangential force is $F_t = 250 \ N$

The angular acceleration is $\alpha = 0.940 \ rad/s^2$

The torque experienced by the wheel is mathematically represented as

$\tau = I * \alpha$

Where I is the moment of inertia

The torque experienced by the wheel can also be mathematically represented as

$\tau = F_t * r$

substituting values

$\tau = 250 * 0.330$

$\tau = 82.5 \ N\cdot m$

So

$82.5 = I * \alpha$

$82.5 = I * 0.940$

So

$I = 87.8 \ kg \cdot m^2$

This moment of inertia can be mathematically evaluated as

$I = \frac{1}{2} * m* r^2$

substituting values

$87.8 = \frac{1}{2} * m* (0.330)^2$

=> $m = 1612.5 \ kg$

5 0

3 years ago

21. An AC generator has an rms output voltage of 274 V at a frequency of 57.0 Hz.

maks197457 [2]

The answers are as follows;

a) the inductive reactance is 322 ohm

b) The maximum voltage is 387.5 V

c) The rms and maximum currents in the inductor are 1.2 A and 0.85 A.

<h3>What is the reactance?</h3>

The reactance is obtained from;

XL = 2πfL

XL = 2 * 3.14 * 57.0 * 0.900

XL = 322 ohm

The maximum voltage is obtained as;

Vo = Vrms * √2

Vo = 274 V * √2

Vo = 387.5 V

Io = Vo/XL

Io = 387.5 V/ 322 ohm

Io = 1.2 A

Irms = 274 V/322 ohm

Irms = 0.85 A

Learn more about inductive reactance:brainly.com/question/17129912

#SPJ1

4 0

2 years ago

A 34.1-mL sample of benzene at 20.8°C was cooled to its melting point, 5.5 °C, and then frozen at 5.5 °C. Calculate the quantity

posledela

Answer:

4541.8 J

Explanation:

First we find the mass of benzene available

mass = density x volume

= 0.867 x 34.1

= 29.5647 g

Then we find the amount of heat transferred by two processes:

heat tranferred = heat lost during temp drop + heat lost during freezing

= mcΔT + mL

= 29.5647 x 1.74 x (20.8 - 5.5) + 29.5647 x 127

= 4541.7883434 J

= 4541.8 J

3 0

3 years ago

Read 2 more answers

A car came to a stop from a speed of 35 m/s in a time of 8.1 seconds. What was the acceleration of the car?

uranmaximum [27]

Simply subtract the two velocities and divide by 8.1,

$\frac{0 - 35}{8.1} = - 4.32$

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey

5 0

3 years ago

Read 2 more answers

When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is

padilas [110]

Answer:

$P_{C} = 3.2\, atm$

Explanation:

Let assume that gases inside bulbs behave as an ideal gas and have the same temperature. Then, conditions of gases before and after valve opened are now modelled:

Bulb A (2 L, 2 atm) - Before opening:

$P_{A} \cdot V_{A} = n_{A} \cdot R_{u} \cdot T$

Bulb B (3 L, 4 atm) - Before opening:

$P_{B} \cdot V_{B} = n_{B} \cdot R_{u} \cdot T$

Bulbs A & B (5 L) - After opening:

$P_{C} \cdot (V_{A} + V_{B}) = (n_{A} + n_{B})\cdot R_{u} \cdot T$

After some algebraic manipulation, a formula for final pressure is derived:

$P_{C} = \frac{P_{A}\cdot V_{A} + P_{B}\cdot V_{B}}{V_{A}+V_{B}}$

And final pressure is obtained:

$P_{C} = \frac{(2\,atm)\cdot (2\,L)+(4\,atm)\cdot(3\,L)}{5\,L}$

$P_{C} = 3.2\, atm$

5 0

3 years ago