a bug is flying with a velocity of 10m/s into a wind at 3m/s what is the resultant velocity of the bug?

A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun

Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100.0}{300}$

$t=0.33\ sec$

Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$s=0+\dfrac{1}{2}\times9.8\times(0.33)^2$

$s=0.53\ m$

Hence, The distance is 0.53 m.

5 0

3 years ago

Twenty is the _________________ of potassium

Mumz [18]

Twenty is the atomic number of potassium.

7 0

4 years ago

Can someone plz help me with this

Elena-2011 [213]

1st Law: Objects that are in motion tend to stay in motion. This motion can change with external forces.

If you were to stop pedaling on bike while in motion, you will notice that you will keep moving. This is because a moving body (you) has inertia. If there wasn't any friction between the tires and the ground, between the axles and wheel, any air resistance, or any other force that acts against you, then you could be coasting indefinitely! 

2nd Law: Force is equal to the mass times acceleration. 

When you pedal, you are applying a force onto the pedal. This force is then translated through tension to apply torque onto the wheel. Turning the wheel will make you accelerate in the lateral direction. 

3rd Law: For every action, there is an equal and opposite reaction. 

Without this, you could pedal and pedal, but you will be not go anywhere! It is essentially the friction between the tires and the ground that propels you forward. If the ground did not apply to the tire the same amount of force that the tire was applying to the ground, the tire would not "catch" and no friction would be applied. And if there was no third law, the weight of you and your bike would "sink" into the ground because the ground would not be applying a normal force back onto you.

hope this helps and if you have any questions just hmu and ask :)

3 0

3 years ago

A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 385 kV. The secondary of

enot [183]

Answer:

a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new output is 86% of the old output

c) The losses in the new line are 74% the losses in the old line.

Explanation:

a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}$

In the old transformer the ratio of voltages was:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{385} =0.03117\\\\n_2=n_1/0.03117=32.1n_1$

In the new transformer the ratio of voltages is:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1$

In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.

b) The new current ratio is

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1$

If the old current output was 425 kV, the ratio of current was:

$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1$

Then, the ratio of the new output over the old output is:

$\frac{I_{2new}}{I_{2old}} =\frac{0.024\cdot I_1}{0.028\cdot I_1}= 0.86$

The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).

c) The power loss is expressed as:

$P_L=I^2\cdot R$

Then, the ratio of losses is (R is constant for both power losses):

$\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74$

The losses in the new line are 74% the losses in the old line.

7 0

3 years ago

pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operat

omeli [17]

Answer:

The maximum permissible propagation delay per flip flop stage is 100 n sec

Explanation:

1024 ripple counter has 10 J-K flip flops(210 = 1024).

So the total delay will be 10×x where x is the delay of each J-K flip flops.

The period of the clock pulse is 1× 10⁻⁶ s.

Now

10x <= 10⁻⁶ s

x <= 100 ns

x= 100 ns for prpoer operation.

pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operation of the counter, the maximum permissible propagation delay per flip flop stage is 100 n sec.

4 0

3 years ago