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3 years ago

What is the work of a 520n student walking a distance of 3 m

Physics

1 answer:

Wewaii [24]3 years ago

8 0

Answer:

W=1560 Joules

Explanation:

Use W=fd

so, W=520N(3m)= 1560J

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valentinak56 [21]

Answer:

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5 0

3 years ago

If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Anon25 [30]

<h2>Answer:50 $ms^{-1}$ </h2>

Explanation:

Let $v_{a}$ be the airspeed.

Let $v_{w}$ be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If $v_{1}$ and $v_{2}$ are two perpendicular vectors,the resultant vector has the magnitude $\sqrt{|v|_{1}^{2}+|v|_{2}^{2}}$

Given,

$v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}$

So,the ground speed is $\sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}$

6 0

3 years ago

Which instrument, a cello or a violin, produces a higher-pitched sound? Explain your answer.

anastassius [24]

I'm pretty sure a violin produces a higher-pitched sound. I play the violin and it's pretty loud. (Hope this helps)

4 0

3 years ago

Read 2 more answers

g A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 ra

Setler [38]

Answer:

The angular acceleration of the wheel is 15.21 rad/s².

Explanation:

Given that,

Time = 5 sec

Final angular velocity = 96.0 rad/s

Angular displacement = 28.0 rev = 175.84 rad

Let $\alpha$ be the angular acceleration

We need to calculate the angular acceleration

Using equation of motion

$\theta=\omega_{i} t+\dfrac{1}{2}\alpha t^2$

Put the value in the equation

$175.84=\omega_{i}\times 5+\dfrac{1}{2}\times\alpha\times(5)^2$

$175.84=\omega_{i}\times 5+12.5\alpha$ ......(I)

Again using equation of motion

$\omega_{f}=\omega_{i}+\alpha t$

Put the value in the equation

$96.0=\omega_{i}+\alpha \times 5$

On multiply by 5 in both sides

$480=\omega_{i}\times 5+\alpha\times 25$ ....(II)

On subtract equation (I) from equation (II)

$480-175.84=\alpha(25-5)$

$304.16=\alpha\times20$

$\alpha=\dfrac{304.16}{20}$

$\alpha=15.21\ rad/s^2$

Hence, The angular acceleration of the wheel is 15.21 rad/s².

5 0

3 years ago

Vector A is in the direction 44.0 degrees clockwise from the y-axis. The x-component of A Ax=-15.0 m. Part A: What is the y-comp

Gnesinka [82]

Answer:

a) Y component of the vector =15.54 m

b) Vector magnitude = 21.6 m

Explanation:

The given vector makes 44 degree angle with Y axis, as given. This is same as 90 -44 = 46 degrees with the horizontal or X axis.

b) X component of the given vector = $A_{x}$ = A cos 46 =15

⇒ A = 15/cos 16 = 21.6 m = Total vector magnitude.

a) Y component of the vector = 21.6 sin 46 = 15.54 m

b) A = 21.6 m

3 0

4 years ago