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Elenna [48]
3 years ago
13

Compute your average velocity in the following two cases: (a) You walk 50.2 m at a speed of 2.21 m/s and then run 50.2 m at a sp

eed of 4.11 m/s along a straight track. (b) You walk for 1.00 min at a speed of 2.21 m/s and then run for 1.16 min at 4.11 m/s along a straight track.
Physics
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

a) 2.87 m/s

b) 3.23 m/s

Explanation:

The avergare velocity can be found dividing the length traveled d by the total time t.

a)

For the first part we easily know the total traveled length which is:

d = 50.2 m + 50.2 m = 100.4 m

The time can be found dividing the distance by the velocity:

t1 = 50.2 m / 2.21 m/s = 22.7149 s

t2 = 50.2 m / 4.11 m/s = 12.2141 s

t = t1 +t2 = 34.9290 s

Therefore, the average velocity is:

v = d/t =2.87 m/s

b)

Here we can easily know the total time:

t = 1 min + 1.16 min = 129.6 s

Now the distance wil be found multiplying each velocity by the time it has travelled:

d1 = 2.21 m/s * 60 s = 132.6 m

d2 = 4.11 m/s *(1.16 * 60 s) = 286.056 m

d = 418.656 m

Therefore, the average velocity is:

v = d/t =3.23 m/s

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Therefore,
Sin ∅2 = n1/n2 *Sin ∅1 = 1.33/1 *Sin 40 = 0.4833=> ∅1 = Sin ^- (0.4833) = 28.9 °

The fisherman the sun at 61.1° (90-∅2) above the horizontal.
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3 years ago
What is the speed of a rocket that travels 8000 meters in 12 seconds?
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If an electronin an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T
Anni [7]

Answer:

Explanation:

Given that,

Force is downward I.e negative y-axis

F = -2 × 10^-14 •j N

Magnetic field is westward, +x direction

B = 8.3 × 10^-2 •i T

Charge of an electron

q = 1.6 × 10^-19C

Velocity and it direction?

Force in a magnetic field is given as

F = q(V×B)

Angle between V and B is 270, check attachment

The cross product of velocity and magnetic field

F =qVB•Sin270

2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2

Then,

v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)

v = 1.51 × 10^6 m/s

Direction of the force

Let x be the direction of v

-F•j = v•x × B•i

From cross product

We know that

i×j = k, j×i = -k

j×k =i, k×j = -i

k×i = j, i×k = -j OR -k×i = -j

Comparing -k×i = -j to given problem

We notice that

-F•j = q ( -V•k × B×i)

So, the direction of V is negative z- direction

V = -1.51 × 10^6 •k m/s

6 0
3 years ago
A 20Kg box is pushed across the floor at a constant velocity with a force of 200N. What is the
Shkiper50 [21]

Answer:

The kinetic coefficient of friction between the box and the floor is 1.020.

Explanation:

Let suppose that box is on horizontal ground. According to the Newton's Laws, an object has a net acceleration of zero when either is at rest or moves at constant velocity. Friction is a reaction to the external force that moves the box. Hence, the equation of equilibrium for the 20-Kg box is:

\Sigma F = F-f = 0 (Eq. 1)

Where:

F - External force, measured in newtons.

f - Friction force, measured in newtons.

If we know that F = 200\,N, then the magnitude of the kinetic friction force is:

f = F

f = 200\,N

In addition, friction force is represented by the following formula:

f = \mu_{k}\cdot m\cdot g (Eq. 2)

Where:

\mu_{k} - Kinetic coefficient of friction, dimensionless.

m - Mass, measured in kilograms.

g - Gravitational acceleration, measured in meters per square second.

Now we clear the kinetic coefficient of friction:

\mu_{k} = \frac{f}{m\cdot g}

If we know that f = 200\,N, m = 20\,kg and g = 9.807\,\frac{m}{s^{2}}, then the kinetic coefficient of friction is:

\mu_{k} = \frac{200\,N}{(20\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\mu_{k} = 1.020

The kinetic coefficient of friction between the box and the floor is 1.020.

4 0
3 years ago
Puffy cumulus clouds, which are made of water droplets, occur at lower altitudes in the atmosphere. Wispy cirrus clouds, which a
xxTIMURxx [149]

Answer:

4083.33 m

Explanation:

T_0 = 24.5°C

\alpha = Constant = \dfrac{6^{\circ}C}{1000\ m}

The equation of temperature at a height is given by

T=T_0-\alpha y

The temperature at which the water will form clouds is 0°C

0=24.5-\dfrac{6}{1000}\times y\\\Rightarrow y=\dfrac{-24.5}{-\dfrac{6}{1000}}\\\Rightarrow y=4083.33\ m

The altitude is 4083.33 m

4 0
3 years ago
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