A wave's frequency is 2Hz and its wavelength is 4 m. What is the wave's speed?

A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?

7nadin3 [17]

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

$v_y=v\ sin(25)$

$v_y=22\ m/s\times\ sin(25)$

$v_y=9.29\ m/s$

Hence, the vertical component of the velocity is 9.29 m/s

3 0

3 years ago

What happens when the voltage increases and the resistance stays the same in a electrical circuit?

Orlov [11]

Answer:

The current in the circuit increases

Explanation:

The ohm's law states that the potential across a circuit is proportional to the current in the circuit.

V ∝ I

Where 'V' is the potential difference across the circuit and 'I' is the current in the circuit.

The proportionality constant present in the equation is the resistance of the circuit. Hence, the equation becomes

V = IR

According to the equation, when V is directly proportional to 'I' where 'R' remains as constant, then the change in 'V is brings change in 'I' to make the equation valid.

So, when there is an increase in the voltage, the current on the circuit increases.

4 0

3 years ago

If the torque required to loosen the nut that is holding a flat tire in a place on a car has a magnitude of 35 N•m what minimum

VashaNatasha [74]

Answer: F = 130 N

Explanation: Solution:

Convert first 27 cm to m.

27 cm x 0.01 m / 1 cm = 0.27 m

Calculate the torque using T = Fd

Derive to find force F

F = T /d

= 35 N.m / 0.27 m

= 130 N

4 0

3 years ago

Read 2 more answers

*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co

sp2606 [1]

Answer: The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

$I = \sqrt{\frac{P}{R} }$

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

$I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }$

Then, simplify the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} }$

Rationalize the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}$

Simplify the numerator by finding its factors:

$I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}$

The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

8 0

3 years ago

To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a

irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

= 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

W = F x S joules

Substituting the values in the above equation

W = 1.5 x 10⁶ x 355 · sin θ

= 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J

6 0

3 years ago