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3 years ago

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A car moving with a velocity 15m/s and accelerating by 5m/s² attempts to reach a car moving with 30 m/s velocity.What distance s

hould it cover to reach the velocity of other car?
A)30 m
B) 90 m
C) 67 m
D) 70m

1 answer:

Firlakuza [10]3 years ago

4 0

I think A is correct

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An air-filled pipe is found to have successive harmonics at 700 Hz , 900 Hz , and 1100 Hz . It is unknown whether harmonics belo

Artyom0805 [142]

Answer:

the length of the pipe is 0.85 m or 85 cm

Explanation:

Given the data in the question;

The successive harmonics are; 700 Hz , 900 Hz , and 1100 H

Now, for a closed pipe,

length of pipe (L) = λ/4

Harmonics; 1x, 3x, 5x, 7x, 9x, 11x

1100Hz - 900Hz = 200Hz

⇒ 2x = 200Hz

x = 100Hz ( fundamental frequency )

λ = V/f = 340 /100 = 3.4 m

Now

Length L = λ / 4

L = 3.4 / 4

L = 0.85 m or 85 cm

Therefore, the length of the pipe is 0.85 m or 85 cm

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3 years ago

A mass m is attached to a string connected to a force sensor on a rotating platform. The platform’s angular velocity, ω, can be

makkiz [27]

Answer:

The mass m is 0.332 kg or 332 gm

Explanation:

Given

The platform is rotating with angular speed , $\omega =8.5\, \frac{rad}{sec}$

Mass m is moving on platform in a circle with radius , $r=0.20\, m$

Force sensor reading to which spring is attached , $F=4.8\, N$

Now for the mass m to move in circle the required centripetal force is given by $F=m\omega ^{2}r$

=> $4.8=m\times 8.5 ^{2}\times 0.20$

$=>m=0.332\, kg$

Thus the mass m is 0.332 kg or 332 gm

7 0

3 years ago

An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination o

Ann [662]

Answer:

θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

v_fly² = v_nort² + v_air²

v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

sin θ = v_air / v_fly

θ = sin⁻¹ (v_air / v_fly)

let's calculate

θ = sin⁻¹ (10/120)

θ = 4.78º

with respect to the vertical or 4.78 to the north-east

5 0

3 years ago

A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car

jolli1 [7]

Answer:

After 15 seconds, the green car will catch up with the blue car

Explanation:

Let the time for the green car to catch up with the blue car be T

When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal

Distance covered by blue car after time T is given by: s = ut + 0.5 at²

Where u = 0, a = 0.2 m/s², t = T

S = 0.5 × 0.2 × T² = 0.1 T²

Velocity of blue car, v = u+ at

v = 0.2T

Distance covered by green car at T is given as: S = Velocity × time

Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)

S = 0.2T (T - 7.5)

S = 0.2 T² - 1.5T

Equating the distance covered by the two cars

0.2T² - 1.5T = 0.1T²

0.1T² - 1.5T = 0

T(0.1T - 1.5) = 0

T = 0 or

T = 1.5/0.1 = 15 secs

Therefore, after 15 seconds, the green car will catch up with the blue car

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3 years ago

g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,

Stolb23 [73]

Answer:

$523269.9\ \text{N/m}$

Explanation:

q = Charge

r = Distance

$q_1=25\ \text{C}$

$r_1=3000\ \text{m}$

$q_2=40\ \text{C}$

$r_2=850\ \text{m}$

The electric field is given by

$E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}$

The electric field at the aircraft is $523269.9\ \text{N/m}$

4 0

3 years ago