What is the speed of a wave on a string with a wavelength of 1.75 m and a frequency of 2.0 Hz

In moving out of a dormitory at the end of the semester, a student does 1.20 x 104 J of work. In the process, his internal energ

Zielflug [23.3K]

Answer:

(a) W=1.20×10⁴J

(b) U= -5.46×10⁴J

(c) Q= -4.26×10⁴J

Explanation:

Given that student does 1.20×10⁴J work

(a) W=1.20×10⁴J

Work done by student,so positive sign

During the process, his internal energy decreases by 5.46×10⁴J.

(b) U= -5.46×10⁴J.

As the Energy decreases therefore negative sign

For (c) Q

We know the formula

$Q=W+U\\Q=1.2*10^{4}+(-5.46*10^{4} )\\ Q=-4.26*10^{4}J$

8 0

3 years ago

A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular

ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

$\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0$

Where,

$\theta =$ Angular Displacement

$\alpha =$ Angular Acceleration

$\omega_0 =$ Angular velocity

$\theta_0 =$ Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

$\theta = \frac{1}{2}\alpha t^2$

Re-arrange for $\alpha,$

$\alpha = \frac{2\theta}{t^2}$

Replacing our values,

$\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}$

$\alpha = 0.139rad/s$

Therefore the ANgular acceleration of the mass is $0.139rad/s^2$

4 0

3 years ago

Which descriptions best fit the labels? X: kinetic energy Y: potential energy X: potential energy Y: kinetic energy X: mechanica

Pani-rosa [81]

W-APE. For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative APE. There must be a minus sign in front of APE to make W positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

( The capital A’s in the words are supposed to be triangles ! I also hoped this helped ! Please mark me as brainliest !! )

3 0

3 years ago

A motorcycle is traveling east at a rate of 56mph.it take 6.7 s for it to decrease its speed to 35mph. How far does the motorcyc

Phantasy [73]

Initial speed = 56mph
Final speed = 35mph

Time taken = 6.7seconds...

Converting the time to hour.. Divide by 3600..
= 6.7/3600

=0.00186hour..

Acceleration = v-u/t

a = 35-56/0.00186
a = -11283.6mph²

The negative sign shows that it decelerated...

V² = u²+2as

(35)² = (56)² + 2×-11283.6×s
Where s is the distance covered within that time...

1225 = 3136 - 22567.2s

22567.2s = 3136-1225

22567.2s = 1911

S = 1911/22567.2

S = 0.08468miles...

But at the end of the question we were made to understand that 1miles = 5280ft

Therefore 0.08468miles = (0.08468×5280)ft

= 447. 11feets...

Which is approximately 447ft.....

Hope this helped.... ?

8 0

3 years ago

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but w

kramer

Answer:

a)

1.35 kg

b)

2.67 ms⁻¹

Explanation:

a)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = ?

$v_{1i}$ = initial velocity of the first body before collision = $v$

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

$v_{1f}$ = final velocity of the first body after collision =

using conservation of momentum equation

$m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}$

Using conservation of kinetic energy

$m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35$

b)

$m_{1}$ = mass of first body = 2.7 kg

$m_{2}$ = mass of second body = 1.35 kg

$v_{1i}$ = initial velocity of the first body before collision = 4 ms⁻¹

$v_{2i}$ = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

$v_{cm} = \frac{(m_{1} v_{1i} + m_{2} v_{2i})}{(m_{1} + m_{2})}\\v_{cm} = \frac{((2.7) (4) + (1.35) (0))}{(2.7 + 1.35)}\\\\v_{cm} = 2.67$ ms⁻¹

8 0

3 years ago