Calculate the current if a charge of 5.60C passes through a point in a conductor in 15.4s

An effort of 200 N is used to lift a load of 800 N by using a lever. If the load is at a distance of 40 cm from the fulcrum then

Alchen [17]

The effort distance will be 160 cm.Applying the moment at the center as follows will provide the effort distance:

<h3 /><h3>What is the mechanical advantage?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtain the efficiency of forces in levers and pulleys.

Given data;

Effort, $\rm F_e=00 N$

Load, $\rm P= 400 \ N$

Distance from the fulcrum, $\rm d=40 \ cm$

The effort distance is found by applying the moment at the center as;

$\rm F_e \times d= P \times d' \\\\ 200 \ N \times d'=800 \ N \times 40 \ cm \\\\ d'=160 \ cm$

Hence, the effort distance will be 160 cm

To learn more about the mechanical advantage refer to the link;

brainly.com/question/7638820

#SPJ1

7 0

2 years ago

Drag each label to the correct location.

34kurt

Electron
3-4-6

proton
1-5-7

neutron
2-5-7

8 0

4 years ago

A certain planet has a radius of 4990 km. If, on the surface of that planet, a 95.0 kg object has a weight of 591 N, then what i

aniked [119]

Answer:

3743.489 kg

Explanation:

F_g = 591 N

G = 6.674x10^-11 constant of gravity

m_1 = 95 kg

m_2 = unknown

r = 4990*1000 =

F_g = G[(m_1*m_2)/r^2]

591 N = 6.674x10^-11[(95*m_2)/4990^2]

8.855 = [(95*m_2)/4990^2]

355631.472 = 95*m_2

m_2 = 3743.489 kg

7 0

3 years ago

A horizontal insulating rod of length 11.8-cm and charge 19 nC is in a plane with a long straight vertical uniform line charge.

SCORPION-xisa [38]

Answer:

11.962337 × 10^-4 N

Explanation:

Given the following :

Length L = 11.8

Charge = 29nC = 29 × 10^-9 C

Linear charge density λ = 1.4 × 10^-7 C/m

Radius (r) = 2cm = 2/100 = 0.02 m

Using the relation:

E = 2kλ/r ; F =qE

F = 2kλq/L × ∫dr/r

F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))

2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4

In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214

Hence,

(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N

3 0

4 years ago

100 g of Ice at -10°C is added into a

Andrei [34K]

Answer:

The mass of the juice responsible for melting the ice is 949.043 grams.

Explanation:

By the First Law of Thermodynamics, we understand that juice releases heat to the ice, which turns into water under the assumption that interactions between the ice-juice system and surroundings are negligible and energy processes are done in steady-state. Since juice is done with water, its specific heat will be taken as of the water. The process is described by the following formula:

$m_{i} \cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})] + m_{w} \cdot c_{w}\cdot (T_{4}-T_{3}) = 0$ (1)

Where:

$m_{i}$ - Mass of ice, in grams.

$m_{w}$ - Mass of the juice, in grams.

$c_{i}$ - Specific heat of ice, in joules per gram-degree Celsius.

$c_{w}$ - Specific heat of water, in joules per gram-degree Celsius.

$L_{f}$ - Latent heat of fusion, in joules per gram.

$T_{1}$ - Initial temperature of ice, in degrees Celsius.

$T_{2}$ - Melting point of water, in degrees Celsius.

$T_{3}$ - Final temperature of the ice-juice system, in degrees Celsius.

$T_{4}$ - Initial temperature of the juice, in degrees Celsius.

If we know that $m_{i} = 100\,g$ , $c_{i} = 2.090\,\frac{J}{g\cdot ^{\circ}C}$ , $c_{w} = 4.18\,\frac{J}{g\cdot ^{\circ}C}$ , $L_{f} = 334\,\frac{J}{g}$ , $T_{1} = -10\,^{\circ}C$ , $T_{2} = 0\,^{\circ}C$ , $T_{3} = 10\,^{\circ}C$ and $T_{4} = 20\,^{\circ}C$ , then the mass of the juice is:

$m_{w} = \frac{m_{i}\cdot [c_{i}\cdot (T_{1}-T_{2}) - L_{f} + c_{w}\cdot (T_{2}-T_{3})]}{c_{w} \cdot (T_{3}-T_{4})}$

$m_{w} = \frac{(100\,g)\cdot \left[\left(2.090\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) - 334\,\frac{J}{g} +\left(4.18\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C) \right]}{\left(4.180\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (-10\,^{\circ}C)}$

$m_{w} = 949.043\,g$

The mass of the juice responsible for melting the ice is 949.043 grams.

5 0

3 years ago