All categories

3 years ago

if the 50 kg object slows down to a velocity of 1 m/s how much kinetic energy does it have 100j 50j 25j or none

Physics

1 answer:

rodikova [14]3 years ago

6 0

$ek = \frac{1}{2}m {v}^{2}$

$ek = \frac{1}{2}(50) {(1)}^{2}$

$ek = \frac{1}{2}(50)$

$ek = 25j$

I'm not really sure but I do know that it's not 0 because the object is still moving, even if it's only moving at 1 m/s.

You might be interested in

Explanation for question i & ii. Thank you.

suter [353]

Do u know Chinese ?if yes I can explain to u easily

6 0

3 years ago

Convert 93.6 miles per hour. Convert this to kilometers per hour.

Readme [11.4K]

Answer:

150.6 km

Explanation:

One mile is about 1.61 km so multiply 93.6 by 1.6 which gives you above 150.6

3 0

3 years ago

You hear a sound with a frequency of 256 Hz. The amplitude of the sound increases and decreases periodically: it takes 2 seconds

german

To solve this problem it is necessary to take into account the concepts related to frequency and period, and how they are related to each other.

The relationship that defines both agreements is given by the equation,

$f_{beat}=\frac{1}{T}$

Then the frequency for the previous period given (2sec) is

$f_{beat}=\frac{1}{2}$

$f_{beat} = 0.5Hz$

The beat frequency of two frequencies is equal to the difference between the two frequencies, then

$f_{beat} = |f_1-f_2|\\f_{beat} = |256Hz-2Hz|\\f_{beat} = 254Hz$

<em>Hence option A is incorrect.</em>

We can do this process for 254Hz as $f_1$ and 258 Hz for $f_2$ , then

$f_{beat} =|254Hz-258Hz|$

$f_{beat} = 4Hz$

<em>Hence option B is incorrect. </em>

We can also do this process for 255Hz as $f_1$ and 257 Hz for $f_2$ , then

$f_{beat} =|255Hz-257Hz|$

$f_{beat} = 2Hz$

<em>Hence option C is incorrect. </em>

We can also do this process for 255.5Hz as f_1 and 256.5 Hz for f_2, then

$f_{beat} =|255.5Hz-256.5Hz|\\f_{beat} = 1Hz$

<em>Hence option D is incorrect. </em>

We can also do this process for 255.75Hz as $f_1$ and 256.25 Hz for $f_2$ , then

$f_{beat} =|255.75Hz-256.25Hz|\\f_{beat} = 0.5Hz$

<em>Hence option E is incorrect. </em>

Therefore the sum of the frequencies in the sound wave would be 256.25Hz and 255.75Hz

3 0

3 years ago

A traffic light is weighing 200N hangs from a vertical cable tied to two other cables that are fastened to a support. The upper

Levart [38]

Answer:

T₁ = 93.6 N , T₂ = 155.6 N , T₃ = 200 N

Explanation:

This is a balance exercise where we must apply the expressions for translational balance in the two axes

∑ F = 0

Suppose that cable t1 goes to the left and the angles are 41º with respect to the horizontal and cable t2 goes to the right with angles of 63º

decompose the tension of the two upper cables

cos 41 = T₁ₓ / T1

sin 41 = T₁y / T1

T₁ₓ = T₁ cos 41

T₁y= T₁ sin 41

for cable gold

cos 63 = T₂ / T₂

sin 63 = $T_{2y}$ / T₂

We apply the two-point equilibrium equation: The junction point of the three cables and the point where the traffic light joins the vertical cable.

Let's start by analyzing the point where the traffic light meets the vertical cable

T₃ - W = 0

T₃ = W

T₃ = 200 N

now let's write the equations for the single point of the three wires

X axis

- T₁ₓ + T₂ₓ = 0

T₁ₓ = T₂ₓ

T1 cos 41 = T2 cos 63

T1 = T2 cos 63 / cos 41 (1)

y Axis

$T_{1y}$ + T_{2y} - T3 = 0

T₁ sin 41 + T₂ sin 63 = T₃ (2)

to solve the system we substitute equation 1 in 2

T₂ cos 63 / cos 41 sin 41 + T₂ sin 63 = W

T₂ (cos 63 tan 41 + sin 63) = W

T₂ = W / (cos 63 tan 41 + sin 63)

We calculate

T₂ = 200 / (cos 63 tan 41 + sin 63)

T₂ = 200 / 1,2856

T₂ = 155.6 N

we substitute in 1

T₁ = T₂ cos 63 / cos 41

T₁ = 155.6 cos63 / cos 41

T₁ = 93.6 N

therefore the tension in each cable is

T₁ = 93.6 N

T₂ = 155.6 N

T₃ = 200 N

6 0

3 years ago

The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave?

Vedmedyk [2.9K]

The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?

<h3>Given:-</h3>

Velocity (V) = 40 m/s

Wavelength $(\lambda)$ = 80 cm = 0.8 m

The frequency (F) of the wave.

<h2>Solution:-</h2>

We know,

$\bf V \: = \: F \: × \: \lambda$

40 = F × 0.8

F = $\frac{40}{0.8}$

F = 50

<h3>The frequency of the wave is <u>5</u><u>0</u><u> </u><u>H</u><u>z</u>. [Answer]</h3>

7 0

3 years ago