1. A 6-volt battery produces a current of 0.5 amps. What is the power in the circuit?

The answer is A and no matter how many times I tried I can't get it.

ryzh [129]

Imagine a skinny straw in the water, standing right over the hole. The WEIGHT of the water in that straw is the force on the tape. Now, the volume of water in the straw is (1 mm^2) times (20 cm). Once you have the volume, you can use the density and gravity to find the weight. And THAT's the force on the tape. If the tape can't hold that force, then it peels off and the water runs out through the hole. /// This is a pretty hard problem, because it involved mm^2, cm, and m^3. You have to be very very very careful with your units as you work through this one. If you've been struggling with it, I'm almost sure the problem is the units.

5 0

2 years ago

What are the different type of insulators in a torch?

Naily [24]

I believe they are:

Pin Insulators
Suspension Insulators
Strain Insulators

If thats not enough theres also insulators commonly used of lower voltage which are:
Stay
Shackle

6 0

2 years ago

(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma

andrew11 [14]

Answer:

a) $F=1.044\times 10^9\ N$

b) $F'=1.044\times 10^9\ N$

c) $F_p=1.0672\times10^{-7}\ N$

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

mass of Mars, $M=6.4\times 10^{23}\ kg$

radius of the Mars, $r=3.4\times 10^{6}\ m$

mass of human, $m=80\ kg$

a)

Gravitation force exerted by the Mars on the human body:

$F=G.\frac{M.m}{r^2}$

where:

$G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2}$ = gravitational constant

$F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}$

$F=1.044\times 10^9\ N$

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

$F'=F$

$F'=1.044\times 10^9\ N$

c)

When a similar person of the same mass is standing at a distance of 4 meters:

$F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}$

$F_p=1.0672\times10^{-7}\ N$

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.

Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.

4 0

2 years ago

I will be so thankful if u answer correctly!!

olga_2 [115]

<h2>Answer:</h2>

(a) 10N

<h2>Explanation:</h2>

The sketch of the two cases has been attached to this response.

<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>

In this case, a frictional force $F_{r}$ is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;

∑F = ma -------------------(i)

Where;

∑F = effective force acting on the object (box)

m = mass of the object

a = acceleration of the object

∑F = F - $F_{r}$

m = 50kg

a = 0 [At constant velocity, acceleration is zero]

<em>Substitute these values into equation (i) as follows;</em>

F - $F_{r}$ = m x a

F - $F_{r}$ = 50 x 0

F - $F_{r}$ = 0

F = $F_{r}$ -------------------(ii)

<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>

In this case, the same frictional force $F_{r}$ is opposing the movement of the box.

∑F = 1.5F - $F_{r}$

m = 50kg

a = 0.1m/s²

<em>Substitute these values into equation (i) as follows;</em>

1.5F - $F_{r}$ = m x a

1.5F - $F_{r}$ = 50 x 0.1

1.5F - $F_{r}$ = 5 ---------------------(iii)

<em>Substitute </em> $F_{r}$ <em> = F from equation (ii) into equation (iii) as follows;</em>

1.5F - F = 5

0.5F = 5

F = 5 / 0.5

F = 10N

Therefore, the value of F is 10N

<em />

4 0

1 year ago

If the velocity versus time graph of an object is a horizontal line, the object is

FromTheMoon [43]

A. Moving with constant non-zero speed

3 0

2 years ago

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