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3 years ago

5. A body travels in a circle at a constant speed, it

2 answers:

7 0

No acceleration as the term acceleration describes change in velocity so if its speed is same there will be no acceleration

nata0808 [166]3 years ago

5 0

A the same acceleration

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An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the

vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = 25.08m/s

Therefore, the initial speed "u" = 25m/s

6 0

3 years ago

A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 16

erastovalidia [21]

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle $\phi$ = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

$V =\pi D( \frac{N}{60} )$

$V = 3.14 (0.034) \frac{(1650)}{60}$

$V = 2.93 \frac{m}{s}$

Form factor for the pinion gear is

Y = 0.303

Now

$K_{v} = \frac{6.1 +0.303}{6.1} = 1.049$

Force on gear tooth

$F = \frac{P}{V}$

$F = \frac{1200}{2.93}$

F = 408.73 N

Now the bending stress is given by the formula

$\sigma = \frac{K_{v} F}{m b y}$

$\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}$

$\sigma$ = 35.38 M pa

This is the value of bending stress on the pinion

8 0

3 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

7 0

3 years ago

1. A large turbine has an initial angular momentum of 6700 kgm^2/s. A storm is rolling in and the wind picks up. 8 seconds later

LekaFEV [45]

Answer:

262.5 Nm

Explanation:

Torque is the rate of change of angular momentum.

Hence, we have

$\tau = \dfrac{\Delta L}{t}$

ΔL is the change in angular momentum.

Using values in the question,

$\tau = \dfrac{8800-6700 \text{ kg m}^2\text{/s}}{8\text{ s}} = 262.5 \text{ Nm}$

4 0

3 years ago

Please answer faast!!!

Drupady [299]

Answer:

Any medium or material of high refractive density e.g Water which refracts the light rays 1&2 away from their normal

7 0

3 years ago