The heart working with the blood vessels to pump blood is which body system?

A gasoline engine receives 200 J of energy from combustion and loses 150 J as heat to the exhaust. What is its efficiency? 75% 2

choli [55]

50/200×100%=25% is answer the formula is usefull energy output divided by total energy provided into 100%

8 0

3 years ago

Read 2 more answers

What happens to the strength of the magnetic field as you come closer to the current carrying wire

Leokris [45]

Form concentric circles around the wire

6 0

4 years ago

If you jog at a speed of 1.5m/s for 20 seconds how far di you travel

Harlamova29_29 [7]

Answer: 30m

Explanation:

Given:

Speed: 1.5m/s

Time: 20 seconds

Distance = speed × time

Distance = 1.5 × 20

= 30m

Therefore you will travel 30m

Must click thanks and mark brainliest

3 0

3 years ago

A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak

GarryVolchara [31]

Explanation:

It is given that,

Mass of the box, m = 100 kg

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.

From the attached figure, the x and y component of forces is given by :

$T_{1x} =-T_1 cos (20)$

$T_{2x} = T_2 cos (40)$

$mg_x = 0$

$T_{1y} = T_1 sin (20)$

$T_{2y} = T_2 sin (40)$

$mg_y= -mg$

Let $R_x$ and $R_y$ is the resultant in x and y direction.

$R_x=-T_1 cos (20)+T_2 cos (40)+0$

$R_y=T_1 sin(20)+T_2 sin(40)-mg$

As the system is balanced the net force acting on it is 0. So,

$-T_1 cos (20)+T_2 cos (40)+0=0$ .............(1)

$T_1 sin(20)+T_2 sin(40)-100\times 9.8=0$ ..................(2)

On solving equation (1) and (2) we get:

$T_1=866.86\ N$ (tension on the left rope)

$T_2=1063.36\ N$ (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.

7 0

3 years ago

Electric fields up to 2.00 × 10 5 N/C have been measured inside of clouds during electrical storms. Neglect the drag force due t

katrin2010 [14]

Answer:

$1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

23.4843749996 m

Yes

Explanation:

E = Electric field = $2\times 10^5\ N/C$

c = Speed of light = $3\times 10^8\ m/s$

m = Mass of proton= $1.67\times 10^{-27}\ kg$

q = Charge of electron = $1.6\times 10^{-19}\ C$

Acceleration is given by

$a=\dfrac{Eq}{m}\\\Rightarrow a=\dfrac{2\times 10^5\times 1.6\times 10^{-19}}{1.67\times 10^{-27}}\\\Rightarrow a=1.9161676647\times 10^{13}\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{1.9161676647\times 10^{13}}{9.81}\\\Rightarrow a=1.9532799844\times 10^{12}g$

The acceleration is $1.9161676647\times 10^{13}\ m/s^2$ or $1.9532799844\times 10^{12}g$

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{(0.1\times 3\times 10^8)^2-0^2}{2\times 1.9161676647\times 10^{13}}\\\Rightarrow s=23.4843749996\ m$

The distance is 23.4843749996 m

The gravitational field is very small compared to the electric field so the effects of gravity can be ignored.

5 0

3 years ago