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2 years ago

10) For a horizontally launched projectile, decreasing the velocity of the

Physics

1 answer:

Daniel [21]2 years ago

4 0

Answer:i think the answer is b

Explanation:because if the velocity is decreased then it will not travel as far and wont stay in the air as long as it wont have as much force keeping it in the air

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A jet can travel the 6000 km distance between washington,

Andrei [34K]

d = r*t is the basic distance equation
d = 6000 km
t with the tail wind = 6 hr
r with the tail wind = speed of the plane + wind speed = s + w
t with the head wind = 7.5 hr
r with the head wind = speed of the plane - wind speed = s-w
(s+w)*6 = 6000
(s-w)*7.5 = 6000
s + w = 1000
s - w = 800
 2s = 1800
s = 900 km/h
s + w = 1000
w = 100
Check the anwer by calculating the return trip.
(900-100) * 7.5 = 800 * 7.5
800 * 7.5 = 6000 km
Answer: The rate of the jet in still air is 900 km/h. The rate of the wind is 100 km/hr.

6 0

3 years ago

An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a w

adell [148]

Answer:

power emitted is 1.75 W

Explanation:

given data

length l = 5 cm = 5 × $10^{-2}$ m

diameter d = 0.074 cm = 74 × $10^{-5}$ m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 × $10^{-5}$ ×5 × $10^{-2}$

area = 1162.3892 × $10^{-5}$ m²

so here power emitted is express as

power emitted = E × σ × area × (temperature)^4

put here all value

power emitted = 0.300× 5.67 × 1162.3892 × $10^{-5}$ × (3068)^4

power emitted = 1.75 W

5 0

3 years ago

A motor running at 2600 rev/min is suddenly switched off and decelerates uniformly to rest after 10 s. Find the angular decelera

Mama L [17]

Angular acceleration = (change in angular speed) / (time for the change)

change in angular speed = (zero - 2,600 RPM) = -2,600 RPM

time for the change = 10 sec

Angular acceleration = -2600 RPM / 10 sec = -260 rev / min-sec

(-260 rev/min-sec) x (1 min / 60 sec) = -(4 1/3) rev / sec²

Since the acceleration is negative, the motor is slowing down.
You might call that a 'deceleration' of (4 1/3) rev/sec² .

The average speed is 1/2(2,600 + 0) = 1,300 rev/min = (21 2/3) rev/sec.

Number of revs = (average speed) x (time) = (21 2/3) x (10sec) = (216 2/3) revs

6 0

3 years ago

Which pair of quantities includes one quantity that increases as the other decreases during simple harmonic motion?

11111nata11111 [884]

Kinetic energy and potential energy pair is the quantity in which one will increase then other will decrease

As we know that sum of kinetic energy and potential energy will always remain conserved

So here we will have

$KE + PE = constant$

so here as we move away from mean position the kinetic energy will decrease while at the same time potential energy will increase.

So the pair of potential energy and kinetic energy will satisfy the above condition

7 0

3 years ago

Read 2 more answers

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0

3 years ago