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3 years ago

How do the forces on a skydiver affect the velocity of the skydiver as they fall? Hellp

2 answers:

8 0

When a skydiver falls, he constantly accelerates. Therby, falling faster and faster. But also he is slowed down a bit due to the force of air resistance.

solmaris [256]3 years ago

3 0

Answer:

Explanation:

As a skydiver falls, he accelerates downwards, gaining speed with each second. The increase in speed is accompanied by an increase in air resistance. This force of air resistance counters the force of gravity.

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Carlos is baking a cake. The last step in the directions is to put the cake batter in the oven. Why does Carlos need to put the

Schach [20]

Answer:

it's d

Explanation:

have you ever baked a cake? Everytime you put cake batter into the oven, it takes a few hours for it to start turning into cake.

4 0

3 years ago

Read 2 more answers

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

morpeh [17]

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron, $u_2=0$

Radius, $r_1=0$

$r_2=2.3 cm=\frac{2.3}{100}=0.023 m$

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron, $m=9.1\times 10^{-31} kg$

Charge on an electron, $q=-1.6\times 10^{-19} C$

Velocity, $v=\frac{Bqr}{m}$

Using the formula

Speed of electron, $v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0$

Speed of second electron, $v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}$

$v_2=1.5\times 10^8 m/s$

Kinetic energy of incident electron= $\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2$

Kinetic energy of incident electron= $0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J$

Kinetic energy of incident electron= $\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV$

1KeV=1000eV

3 0

3 years ago

A block of weight mg sits on an inclined plane as shown in (Figure 1) . A force of magnitude F1 is applied to pull the block up

svlad2 [7]

Answer:

W = (F1 - mg sin θ) L, W = -μ mg cos θ L

Explanation:

Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane

Y Axis

N - $W_{y}$ =

N = W_{y}

X axis

F1 - fr - Wₓ = 0

fr = F1 - Wₓ

Let's use trigonometry to find the components of the weight

sin θ = Wₓ / W

cos θ = W_{y} / W

Wₓ = W sin θ

W_{y} = W cos θ

We substitute

fr = F1 - W sin θ

Work is defined by

W = F .dx

W = F dx cos θ

The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1

W = -fr x

W = (F1 - mg sin θ) L

Another way to calculate is

fr = μ N

fr = μ W cos θ

the work is

W = -μ mg cos θ L

4 0

3 years ago

___________ force can be thought of as either a push or a pull, as long as it makes the object move in a circular path.

fomenos

I believe it would be the Centripetal force.

5 0

3 years ago

Read 2 more answers

It decomposes in a first-order reaction with a rate constant of 14 s–1. how long would it take for an initial concentration of 0

VikaD [51]

The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have

$rate = \frac{-(C_{new} - C_{old})}{t}$

where C are the concentration values and t is the time taken for it to decompose.

$14 = \frac{-(0.02 - 0.06)}{t}$
$t = 0.003 s$

Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s

4 0

3 years ago