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topjm [15]
2 years ago
13

* 1 po What are the health benefits of having muscular strength and endurance?

Physics
1 answer:
CaHeK987 [17]2 years ago
3 0

Answer:

i can`t help without showing me the choses

Explanation:

You might be interested in
Estimate the inductance L of a coil that is 12 cm long, made of about 235 copper-wire turns and a diameter of about 1.7 cm. Show
ANTONII [103]

Answer:

Inductance as calculated is 13.12 mH

Solution:

As per the question:

Length of the coil, l = 12 cm = 0.12 m

Diameter, d = 1.7 cm = 0.017 m

No. of turns, N = 235

Now,

Area of cross-section of the wire, A = \frac{\pi d^{2}}{4} = \frac{\pi \times 0.017^{2}}{4} = 2.269\times 10^{- 4}\ m^{2}

We know that the inductance of the coil is given by the formula:

L = \frac{mu_{o}AN^{2}}{l} = \frac{4\pi \times 10^{- 7}\times 2.269\times 10^{- 4}\times 235^{2}}{0.12} = 1.312\times 10^{- 4}\ H = 13.12\ mH

4 0
3 years ago
A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.
dalvyx [7]

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

6 0
3 years ago
A 21.0 kg shopping cart is moving with a velocity of 6.0 m/s. It strikes a 11.0 kg box that is initially at rest. They stick tog
DedPeter [7]

Answer:

a) 126 kgm/s

b) 0 kgm/s

c) 3.9 m/s

Explanation:

To solve this question, we will use the law of conservation of momentum.

Momentum before collision = momentum after collision

m1v1 + m2v2 = (m1 + m2)v, where

m1 = mass of the shopping cart, 21 kg

m2 = mass of the box, 11 kg

v1 = initial velocity of the shopping cart, 6 m/s

v2 = initial velocity of the box, 0 m/s

v = final velocity of the box+cart

a)

Momentum of the shopping cart before collision = P

P = mv

P = 21 * 6

P = 126 kgm/s = c

b)

Momentum of the box before collision

Like in question a above, the momentum of the box is P

P = mv

P = 11 * 0

P = 0 kgm/s = b

c)

Velocity of the combined shopping cart wreckage after collision is

m1v1 + m2v2 = (m1 + m2)v

(21 * 6) + (11 * 0) = (21 + 11)v

126 + 0 = 32v

32v = 126

v = 126/32

v = 3.9375 m/s, on approximating to 1 decimal place, we have 3.9 m/s and option b as the answer.

Please vote brainliest

8 0
3 years ago
The leading cause of air pollution is?
stellarik [79]
Combustion Engine Exhaust----Cars, trucks, jet airplanes and other combustion engine vehicles cause air pollution. The exhaust from these contains carbon monoxide, nitrous oxide and gaseous oxide. This type of air pollution creates smog
3 0
3 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
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