All categories

4 years ago

What simple machine can be described as a shaft that is attached to the center of a wheel?

Physics

1 answer:

charle [14.2K]4 years ago

4 0

"Wheel & Axle" <span>can be described as a shaft that is attached to the center of a wheel

Hope this helps!</span>

You might be interested in

A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is def

Sladkaya [172]

Answer:

Speed =0.283m/ s

Direction = 47.86°

Explanation:

Since it is a two dimensional momentum question with pucks having the same mass, we derive the momentum in xy plane

MU1 =MU2cos38 + MV2cos y ...x plane

0 = MU2sin38 - MV2sin y .....y plane

Where M= mass of puck, U1 = initial velocity of puck 1=0.46, U2 = final velocity of puck 1 =0.34, V2 = final velocity of puck 2, y= angular direction of puck2

Substitute into equation above

.46 = .34cos38 + V2cos y ...equ1

.34sin38 = V2sin y...equ2

.19=V2cos Y...x

.21=V2sin Y ...y

From x

V2 =0.19/cost

Sub V2 into y

0.21 = 0.19(Sin y/cos y)

1.1052 = tan y

y = 47.86°

Sub Y in to x plane equ

.19 = V2 cos 47.86°

V2=0.283m/s

7 0

3 years ago

How could physics be useful in weather prediction?

bogdanovich [222]

Wind speed, temperature

4 0

4 years ago

(a) Two identical open boxes originally contain the same volume of water. One is kept at 15°C and the other at 85°C for the same

Tresset [83]

Explanation:

because as you can see beaker 2 has higher temperature. As volume is directly propotional

to temperature. So the the water molecules has

more energy so that's why levels are different.

6 0

2 years ago

Any five physics problems

kow [346]

Explanation:

There are still some questions beyond the Standard Model of physics, such as the strong CP problem, neutrino mass, matter–antimatter asymmetry, and the nature of dark matter and dark energy.

7 0

3 years ago

A blue car with mass mc = 459.0 kg is moving east with a speed of vc = 19.0 m/s and collides with a purple truck with mass mt =

ivolga24 [154]

Answer:

1) $P_c$ = 8721 Kg*m/s

2) $P_t$ = 12450 Kg*m/s

3) θ = 55.03°

4) P = 15182 Kg*m/s

5) V = 8.91 m/s

6) $E_i>E_f$

145099.5 J > 67638 J

Explanation:

1) The momentum is calculated by the next equation:

P = MV

where M is the mass and V is the velocity

so, the linear momentum of the car is:

$P_c$ = (459)(19)

$P_c$ = 8721 Kg*m/s

2) the linear momentum of the truck is:

$P_t$ = (1245)(10)

$P_t$ = 12450 Kg*m/s

3) For answer this we will use the law of the conservation of the linear momentum where:

$P_i = P_f$ :

So, we will do this for each axis:

First on axis x:

$M_cV_c = V_{sx}M_s$

Where $M_c$ is the mass of the blue car, $V_c$ is the velocity of the car, $V_{sx}$ is the velocity of the system in x after the collition and $M_s$ is the mass of both cars. Replacing, we get:

$(459)(19) = V_{sx}(1245+459)$

Solving for $V_{sx}$ :

$V_{sx} = 5.11 m/s$

Second, on axis y:

$M_tV_t = V_{sy}M_s$

Where $M_t$ is the mass of the truck, $V_t$ is the velocity of the truck, $V_{sy}$ is the velocity of the system in y after the collition and $M_s$ is the mass of both cars.