When an atom that has no charge loses two electrons, it becomes a

10 PTS. If an electromagnetic wave has a frequency of 8 ⋅ 10^14 H z , what is its wavelength? Use λ=V/F. The speed of light is 3

ad-work [718]

Answer :

The wavelength would be 410 m/s

3 0

3 years ago

Jeffery gains super strength and pushes two different objects with the same amount of force. Object A accelerates at 40 m/s2, an

Montano1993 [528]

Answer: Object B

Explanation: Acceleration is directly proportional to force and inversely proportional to mass. It implies that more massive objects accelerates at a slower rate.

5 0

3 years ago

A 60kg bicyclist (including the bicycle) is pedaling to the

Fittoniya [83]

a) 4 forces

b) 186 N

c) 246 N

Explanation:

a)

Let's count the forces acting on the bicylist:

1) Weight ( $W=mg$ ): this is the gravitational force exerted on the bicyclist by the Earth, which pulls the bicyclist towards the Earth's centre; so, this force acts downward (m = mass of the bicyclist, g = acceleration due to gravity)

2) Normal reaction (N): this is the reaction force exerted by the road on the bicyclist. This force acts vertically upward, and it balances the weight, so its magnitude is equal to the weight of the bicyclist, and its direction is opposite

3) Applied force ( $F_A$ ): this is the force exerted by the bicylicist to push the bike forward. Its direction is forward

4) Air drag ( $R$ ): this is the force exerted by the air on the bicyclist and resisting the motion of the bike; its direction is opposite to the motion of the bike, so it is in the backward direction

So, we have 4 forces in total.

b)

Here we can find the net force on the bicyclist by using Newton's second law of motion, which states that the net force acting on a body is equal to the product between the mass of the body and its acceleration:

$F_{net}=ma$

where

$F_{net}$ is the net force

m is the mass of the body

a is its acceleration

In this problem we have:

m = 60 kg is the mass of the bicyclist

$a=3.1 m/s^2$ is its acceleration

Substituting, we find the net force on the bicyclist:

$F_{net}=(60)(3.1)=186 N$

c)

We can write the net force acting on the bicyclist in the horizontal direction as the resultant of the two forces acting along this direction, so:

$F_{net}=F_a-R$

where:

$F_{net}$ is the net force

$F_a$ is the applied force (forward)

$R$ is the air drag (backward)

In this problem we have:

$F_{net}=186 N$ is the net force (found in part b)

$R=60 N$ is the magnitude of the air drag

Solving for $F_a$ , we find the force produced by the bicyclist while pedaling:

$F_a=F_{net}+R=186+60=246 N$

3 0

3 years ago

The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.

inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

E = $\frac{\sigma}{2 \times \epsilon_{o}}$

= $\frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$

= $1.18 \times 10^{5} N/C$

Now, we will consider that equilibrium of forces are present there. So,

ma = qE

a = $\frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}$

= $1.13 \times 10^{13} m/s^2$

According to the third equation of motion,

$v^{2} = 2 \times a \times d$

or, d = $\frac{v^{2}}{2d}$

= $\frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}$

= 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

7 0

3 years ago

Find the speed vfinal of the joined cars after the collision. mastering physics

Tanya [424]

<span>Px = 0 Py = 2mV second, Px = mVcosĎ† Py = â€“mVsinĎ† add the components Rx = mVcosĎ† Ry = 2mV â€“ mVsinĎ† Magnitude of R = âš(RxÂ˛ + RyÂ˛) = âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) and speed is R/3m = (1/3m)âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) simplifying Vf = (1/3m)âš((mVcosĎ†)Â˛ + (2mV â€“ mVsinĎ†)Â˛) Vf = (1/3)âš((VcosĎ†)Â˛ + (2V â€“ VsinĎ†)Â˛) Vf = (V/3)âš((cosĎ†)Â˛ + (2 â€“ sinĎ†)Â˛) Vf = (V/3)âš((cosÂ˛Ď†) + (4 â€“ 2sinĎ† + sinÂ˛Ď†)) Vf = (V/3)âš(cosÂ˛Ď†) + (4 â€“ 2sinĎ† + sinÂ˛Ď†)) using the identity sinÂ˛(Ď)+cosÂ˛(Ď) = 1 Vf = (V/3)âš1 + 4 â€“ 2sinĎ†) Vf = (V/3)âš(5 â€“ 2sinĎ†)</span>

6 0

4 years ago