A bicycle consists of which types of simple machines? Check all that apply. lever

Answers

Answer 1

Answer:

Lever, pulley and wheel and axle are the types of machine it's grouped to

Explanation:

The wheel and axle is a simple machine that works by reducing friction in trying to move a load. This is seen in the Tyre of the bicycle

Pulley is a simple machine that creates a mechanical advantage and supports the changing of direction for a rope or cable. This is seen in the chain of the bicycle

Levers attached to the bike's pedals are pushed down to direct force into the pulley system.

Answer 2

Answer: A bike is compound machine.This means that it is made up of a bunch of different simple machines. Screws are used to hold the pieces together. The wheel of a bicycle is a wheel and axle. The pedals are also examples of this simple machine. The pedals are attached to a lever that turns a pulley.

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Consider two copper wires of equal cross-sectional area. One wire has 3 times the length of the other. How do the resistivities of these two wires compare?

Answers

Explanation:

The relation between resistance and resistivity is given by :

$R=\rho (l)/(A)$

$\rho$ is resistivity of material

l is length of wire

A is area of cross section of wire

Resistivity of a material is the hidden property. If one wire has 3 times the length of the other, then it doesn't affect its resistivity. Hence, the resistivity of two wires is

the period of the satellite is exact 42.391 hours, the earth's mass is 5.98 kg and the radius of th earth is 3958.8 miles, what is the distance of the satellite from the surface of the earth in miles?

Answers

Answer:

As the mass is not written well, i will use the equation in terms of the gravitational acceleration:

The equation for the period of a satellite is:

$T = 2*pi*\sqrt{(r^3)/(g*R^2) }$

We want to find r, so isolating r we get:

$r = \sqrt[3]{x ((T)/(2*pi) )^2*g*R^2}$

Where:

T = period.

r = radius of the satellite.

R = radius of the planet.

g = gravitational acceleration of the planet.

pi = 3.14159...

g = 78999.64 mi/h^2 (value of a table)

T = 42.391 h.

R = 3958.8 miles

We can replace those values in the equation and get:

$r = \sqrt[3]{ ((42.391)/(2*3.14159) )^2*78999.64*(3958.8)^2} = 38,339.5 mi$

Now this value is measured from the center of the Earth, then the altitude of the satellite measured from the surface of the Earth will be:

H = r - R = 38,339.1mi - 3958.8mi = 34,380.3 mi

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second particle is at (−7 + 2t, −6 + 2t, −1 + t). (a) (5 Points) Do the paths of the two particles cross? If so, where?

Answers

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .

Setting the locations equal, we get the following equations to solve for $T_F$ and $T_S$ :

$(-1 + T_F) = (-7 + 2T_S)$ Equation 1

$(4 - T_F) = (-6 + 2T_S)$ Equation 2

$(-1 + 2T_F) = (-1 + T_S)$ Equation 3

Solving these three equations simultaneously we get:

$T_F =$ 2 seconds

$T_S =$ 4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of $T_F$ or $T_S$ in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

When the frequency of an electromagnetic wave increases, its energy A. increases.
B. decreases.
C. stays the same.
D. It depends on the exact type of electromagnetic wave.

Answers

At the time when the frequency of an electromagnetic wave increases, its energy increased.

The following information should be considered:

When the electromagnetic wave increases so the energy should be increased.
Due to all this, the speed, wavelength all should be decreased.

Therefore we can conclude that At the time when the frequency of an electromagnetic wave increases, its energy increased.

Learn more: brainly.com/question/6201432

When the frequency of an electromagnetic wave increases, its energy also must increase. As this occurs, its speed, wavelength, and amplitude all decreases. So, the correct choice would be A. Increases.

I hope I was able to satisfyingly answer your question. :)

Air trapped in the cooling system could create undesirable areas of combustion heat buildup in the engine called _____.

Answers

Answer:

Air pockets.

Explanation:

Air pockets in the cooling system are bubbles of air trapped within the lines (hoses and pipes) of the cooling system. This air bubbles enter the cooling system usually during the process of filling the radiator coolant fluid (usually water), or replacing the water pump or the radiator hose during repairs or servicing of the cooling system. The trapped air prevent pressure movement that is needed by the coolant to move the heat generated from the engine cylinder, resulting in heat build up. The solution is to "bleed" the engine through the radiator lid or some air release valves.

A boat that travels 3.00 m/s relative to the water is crossing a river that is 1.00 km wide. The destination on the far side of the river is 0.500 km downstream from the starting point. (a) If the river current is 2.00 m/s, in what direction should the boat be pointed in order to reach the destination? (b) How much time will the trip take?

Answers

Answer:

a) 10.29° upstream

b) t=338.7s

Explanation:

If the river is 1km wide and the destination point is 0.5km away downstream, then the angle and distance the the boat has to travel is:

$\alpha =atan((0.5)/(1))=26.56°$

$D=√(1^2+0.5^2)=1.118km$

The realitve velocity of the boat respect to the water is:

$V_(B/W)=[3*cos\beta ,3*sin\beta ]$ where β is the angle it has to be pointed at.

From the relative mvement equations:

$V_(B/W)=V_B-V_W$ where $V_B=[V*cos\alpha ,-V*sin\alpha ]$

From this equation we get one equation per the x-axis and another for the y-axis. If we square each of them and add them together, we will get 2 equations:

$(3*cos\beta )^2+(3*sin\beta )^2=(V*cos\alpha )^2+(-V*sin\alpha +2)^2$

$V^2-4*V*sin\alpha -5=0$ Solving for V:

V = 3.3m/s and V=-1.514m/s Replacing this value into one of our previous x or y-axis equations:

$\beta =acos((V*cos\alpha )/(3) ) = 10.29°$

The amount of time:

$t = D/V = (1118m)/(3.3m/s) =338.7s$

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