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Anestetic [448]

3 years ago

6

The diagram below a free body diagram picturing a cyclist moving along a highway. 4 forces represented by arrows A,B,C, and D ar

e acting on the cyclist as she pedals. The arrow that would best represent the normal force is:

1 answer:

vlabodo [156]3 years ago

3 0

A because the normal force always acts perpendicular and away from the surface an object is in contact with.

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A student makes the following claim about electron flow through a circuit that includes a battery and a bulb:

nata0808 [166]

Where are all the other claims at?

7 0

3 years ago

A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?

garri49 [273]

<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>

6 0

4 years ago

A floating object oscillates up and down 2 complete cycles in 1 second as a water wave of wavelength 5 meters passes by. The spe

Elena-2011 [213]

Answer:

10 m/s

Explanation:

The speed of a wave is given by

$v=\lambda f$

where

$\lambda$ is the wavelength

f is the frequency

For the wave in this problem,

- The frequency is given by the number of oscillations per second, so

$f=\frac{2 cycles}{1 s}=2 Hz$

- The wavelength is

$\lambda=5 m$

So, the wave speed is

$v=(5 m)(2 Hz)=10 m/s$

3 0

3 years ago

A 10kg box accelerates forward at a rate of 12 m/s^2. What is the force acting on the box?

Allushta [10]

Answer:

120N

Explanation:

Newton's second law formula: F= ma

given that m = 10 kg, a = 12 m/s^2

F = ma = 10 kg * 12 m/s^2 = 120 kgm/s^2 = 120 N

5 0

3 years ago

In unit-vector notation, what is the torque about the origin on a particle located at coordinates (0 m, −3.0 m, 2.0 m) due to fo

irinina [24]

Answer:

The torque about the origin is $2.0Nm\hat{i}-8.0Nm\hat{j}-12.0Nm\hat{k}$

Explanation:

Torque $\overrightarrow{\tau}$ is the cross product between force $\overrightarrow{F}$ and vector position $\overrightarrow{r}$ respect a fixed point (in our case the origin):

$\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}$

There are multiple ways to calculate a cross product but we're going to use most common method, finding the determinant of the matrix:

$\overrightarrow{r}\times\overrightarrow{F} =-\left[\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k}\\ F1_{x} & F1_{y} & F1_{z}\\ r_{x} & r_{y} & r_{z}\end{array}\right]$

$\overrightarrow{r}\times\overrightarrow{F} =-((F1_{y}r_{z}-F1_{z}r_{y})\hat{i}-(F1_{x}r_{z}-F1_{z}r_{x})\hat{j}+(F1_{x}r_{y}-F1_{y}r_{x})\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F} =-((0(2.0m)-0(-3.0m))\hat{i}-((4.0N)(2.0m)-(0)(0))\hat{j}+((4.0N)(-3.0m)-0(0))\hat{k})$

$\overrightarrow{r}\times\overrightarrow{F}=-2.0Nm\hat{i}+8.0Nm\hat{j}+12.0Nm\hat{k}=\overrightarrow{\tau}$

4 0

3 years ago