How does the electric potential energy change as the electron moves from i to f?

A child is riding a bike at a speed of 6m/s with a total kinetic energy of 1224J. If the mass of the child is 30kg, what is the

UkoKoshka [18]

Answer:

Mass of bike = 38 kg.

Explanation:

Kinetic energy is given by the expression, $KE = \frac{1}{2} mv^2$ , where m is the mass and v is the velocity.

Here speed of child riding bike = 6 m/s

Mass of child = 30 kg

Total kinetic energy = 1224 J

Let the mass of bike be, m kg

So, total mass of child and bike = (m + 30) kg

Substituting,

$1224 = \frac{1}{2}* (m+30)*6^2\\ \\ m+30=68\\ \\ m=38kg$

So, mass of bike = 38 kg.

3 0

4 years ago

If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the

erastova [34]

Either 175 N or 157 N depending upon how the value of 48Â° was measured from.
You didn't mention if the angle of 48Â° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 NÂ·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 NÂ·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0Â° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0Â° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48Â° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48Â° is from the lug wrench itself, that means that the force is 90Â° - 48Â° = 42Â° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.

6 0

3 years ago

Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of 1800 rpm. (Round

erma4kov [3.2K]

Answer:

$\tau=3.96\ ksi$

Explanation:

Given that

d= 1.5 in ( 1 in = 0.0254 m)

d= 0.0381 m

P= 75 hp ( 1 hp = 745.7 W)

P= 55927.5 W

N= 1800 rpm

We know that power P is given as

$P=\dfrac{2\pi N\ T}{60}$

T=Torque

N=Speed

$55927.5=\dfrac{2\times \pi \times 1800\ T}{60}$

T=296.85 N.m

The maximum shear stress is given as

$\tau=\dfrac{16 T}{\pi d^3}$

$\tau=\dfrac{16\times 296.85}{\pi \times 0.0381^3}$

$\tau=27.35\ MPa$

We know that 1 MPa =0.145 ksi

$\tau=3.96\ ksi$

3 0

4 years ago

Semi-conductors are materials that

Svetach [21]

Good morning, Jesusperez7!

Semi-conductors are materials that have insulator and conductor properties.

Thank you for choosing Brainly for your tutoring site, and I hope you enjoy your experience here!

6 0

3 years ago

Hello guys! Can u please help me with physics. I translated it in English. Can yall help me please how much u can!!

DedPeter [7]

1. Since the body is thrown vertically upward, the only force acting on it as it rises and falls is gravity, which causes a constant downward acceleration with magnitude g = 9.8 m/s². Because this acceleration is constant, we can use the formula

v² - u² = 2a ∆x

where

u = initial speed

v = final speed

a = acceleration

∆x = displacement

At its maximum height, some distance y above the point where the body is launched, the body has zero velocity, so

0² - (20 m/s)² = 2 (-9.8 m/s²) y

Solve for y :

y = (20 m/s)² / (2 (9.8 m/s²)) ≈ 20.4 m

2. Relative to the ground, the body's maximum height is 60 m + 20.4 m ≈ 80.4 m.

3. At any time t ≥ 0, the body's vertical velocity is given by

v = 20 m/s - gt

At the highest point, we have

0 = 20 m/s - (9.8 m/s²) t

and solving for t gives

t = (20 m/s) / (9.8 m/s²) ≈ 2.04 s

4. The body's height y above the ground at any time t ≥ 0 is given by

y = 60 m + (20 m/s) t - 1/2 gt²

Solve for t when y = 0 :

0 = 60 m + (20 m/s) t - 1/2 (9.8 m/s²) t²

Using the quadratic formula,

t = (-b + √(b² - 4ac)) / (2a)

(and omitting the negative root, which gives a negative solution) where a = -1/2 (9.8 m/s²), b = 20 m/s, and c = 60 m. You should end up with

t ≈ 6.09 s

5. At the time found in (4), the body's velocity is

v = 20 m/s - g (6.09 s) ≈ -39.7 m/s

Speed is the magnitude of velocity, so the speed in question is 39.7 m/s.

6 0

3 years ago