A reaction that results in the combining of smaller atomic nuclei is ___

If you are in an elevator that speed up then the apparent weight is

Sholpan [36]

Answer:

Fnet - Fg

Explanation:

When an object is in an elevator, its weight varies with respect to the direction of movement of the elevator and the elevators acceleration.

The weight, W, of an object can be expressed as;

W = mg

where m is the object's mass, and g is the acceleration due gravity.

If the object is in an elevator that speed up, an apparent weight would be felt since both mass and elevator are moving against gravitational pull of the earth.

So that,

$W_{net}$ = mg + ma

where: mg is the weight of the object, and ma is the apparent weight.

Apparent weight (ma) = $W_{net}$ - mg

3 0

2 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

3 0

2 years ago

Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk.

stellarik [79]

Answer: The correct option is Option b.

Explanation:

Power is defined as the rate of work done by an object.

Mathematically,

$P=\frac{W}{t}$ .....(1)

And work done is the product of force exerted on the object times the displacement covered by that object.

Mathematically,

$W=F.s$

Putting this value in above equation, we get:

$P=\frac{F.s}{t}$

where,

P = power = ?W

F = Force exerted = 10N

s = Displacement = 400cm = 4m (Conversion factor: 1m = 100 cm)

t = Time taken = 8s

Putting values in above equation, we get

$P=\frac{10\times 4}{8}\\\\P=5W$

Hence, the correct option is Option b.

7 0

3 years ago

Read 2 more answers

Helppppppppppp im dieing

Otrada [13]

It would be 1. B 2. A 3. A

7 0

3 years ago

A pole AB of length 10.0m and weight 600N has its center of gravity 4.0m from the end A, and lies on horizontal ground .Calculat

postnew [5]

Answer:

The force required to begin to lift the pole from the end 'A' is 240 N

Explanation:

The given parameters for the pole AB are;

The length of the pole, l = 10.0 m

The weight of the pole, W = 600 N ↓

The distance of the center of gravity of the pole from the side 'A' = 4.0 m

Let ' $F_A$ ' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive

For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have

$F_A$ × 10.0 m - W × 4.0 m = 0

∴ $F_A$ × 10.0 m = W × 4.0 m = 600 N × 4.0 m

$F_A$ × 10.0 m = 600 N × 4.0 m

∴ $F_A$ = 600 N × 4.0 m/(10.0 m) = 240 N

The force required to begin to lift the pole from the end 'A', $F_A$ = 240 N.

8 0

2 years ago

A reaction that results in the combining of smaller atomic nuclei is ____.