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Gekata [30.6K]
3 years ago
8

A ceiling fan with 90-cm-diameter blades is turning at 64 rpm . Suppose the fan coasts to a stop 28 s after being turned off. Wh

at is the speed of the tip of a blade 10 s after the fan is turned off
Physics
1 answer:
mixas84 [53]3 years ago
4 0

Answer:

the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

Explanation:

Given;

diameter of the ceiling fan, d = 90 cm = 0.9 m

angular speed of the fan, ω = 64 rpm

time taken for the fan to stop, t = 28 s

The distance traveled by the ceiling fan when it comes to a stop is calculated as;

d = vt = \omega r\times  t= ( \frac{64 \ rev}{\min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \min}{60 \ s} \times 0.9 \ m) \times 28 \ s\\\\d = 168.89 \ m

The speed of the tip of a blade 10 s after the fan is turned off is calculated as;

v = \frac{d}{t} \\\\v = \frac{168.89}{10} \\\\v = 16.889 \ m/s

Therefore, the speed of the tip of a blade 10 s after the fan is turned off is 16.889 m/s.

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A car travelling at 15 m/s comes to rest in a distance of 14 m when the brakes are applied.
stira [4]

Answer:

-8.04 m/s2

Explanation:

To find the answer to this, you have to use the 4th kinematic equation:

v^{2} = v^{2}_{0}  + 2ax

You plug into the equation to get:

0 = 15^{2} + 2a(14)

solve for a to get

-8.04 m/s2

3 0
3 years ago
A needle 5cm long can just rest on the surface of the water without wetting. What us it's weight? Surface tension is 0.007N/m​
yulyashka [42]

Answer:

c

Explanation:

6 0
3 years ago
Samples of different materials, A and B, have the same mass, but the sample
Effectus [21]

Answer:

B. The particles that make up material B have more mass than the

particles that make up material A.

Explanation:

3 0
3 years ago
A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0
3 years ago
The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Julli [10]

Answer:

v=(6ti+6k)\ m/s

Explanation:

Given that,

The position of a particle is given by :

r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m

Let us assume we need to find its velocity.

We know that,

v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s

So, the velocity of the particle is (6ti+6k)\ m/s.

5 0
3 years ago
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