Ask question

Ask question

All categories

1 year ago

9

Calculate the first and second velocities of the car with three washers attached to the pulley, using the formulas v1 = 0.25 m /

t1, and v2 = 0.25 m / (t2 – t1) where t1 and t2 are the average times the car took to reach the 0.25 and the 0.50 meter marks. Record these velocities, to two decimal places, in Table E. What is the first velocity of the car with three washers at the 0.25 meter mark? m/s What is the second velocity of the car with three washers at the 0.50 meter mark? m/s

2 answers:

Ber [7]1 year ago

5 0

Answer:

The 1st one is 0.19

The 2nd one is 0.45

It's right for both of them :)

Explanation:

My name is Ann [436]1 year ago

5 0

Answer:

.25= .13 m/s

.50= .36 m/s

Explanation:

You might be interested in

An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .

miss Akunina [59]

Answer:

The airplane should release the parcel $6.7*10^2$ m before reaching the island

Explanation:

The height of the plane is $y_0=100m$ , and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

$y=y_0 - \frac{gt^2}{2}$ [1]

And the distance X is

x = V.t [2]

Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

$X = V . \sqrt{\frac{2y_0}{g}}$

Using the given values:

$x = 150 m/s \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}$

x = $6.7*10^2$ m

4 0

1 year ago

The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from

Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

$L_T=L_S+10log(n)$

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

$L_T=60+10 log(5)$

= 66.98 db

7 0

1 year ago

A 0.300kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.800m/s when it makes a head-o

e-lub [12.9K]

Answer:

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

0.010935 J

0.0858675 J

Explanation:

$m_1$ = Mass of first glider = 0.3 kg

$m_2$ = Mass of second glider = 0.15 kg

$u_1$ = Initial Velocity of first glider = 0.8 m/s

$u_2$ = Initial Velocity of second glider = 0 m/s

$v_1$ = Final Velocity of first glider

$v_2$ = Final Velocity of second glider

As momentum and Energy is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

${\tfrac {1}{2}}m_{1}u_{1}^{2}+{\tfrac {1}{2}}m_{2}u_{2}^{2}={\tfrac {1}{2}}m_{1}v_{1}^{2}+{\tfrac {1}{2}}m_{2}v_{2}^{2}$

From the two equations we get

$v_{1}=\frac{m_1-m_2}{m_1+m_2}u_{1}+\frac{2m_2}{m_1+m_2}u_2\\\Rightarrow v_1=\frac{0.3-0.15}{0.3+0.15}\times 0.8+\frac{2\times 0.15}{0.3+0.15}\times 0\\\Rightarrow v_1=0.27\ m/s$

The final velocity of the first glider is 0.27 m/s in the same direction as the first glider

$v_{2}=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 0.3}{0.3+0.15}\times 0.8+\frac{0.3-0.15}{0.3+0.15}\times 0\\\Rightarrow v_2=1.067\ m/s$

The final velocity of the second glider is 1.07 m/s in the same direction as the first glider.

Kinetic energy is given by

$K=\frac{1}{2}m_1v_1^2\\\Rightarrow K=\frac{1}{2}0.3\times 0.27^2\\\Rightarrow K=0.010935\ J$

Final kinetic energy of first glider is 0.010935 J

$K=\frac{1}{2}m_2v_2^2\\\Rightarrow K=\frac{1}{2}0.15\times 1.07^2\\\Rightarrow K=0.0858675\ J$

Final kinetic energy of second glider is 0.0858675 J

6 0

2 years ago

1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational

sattari [20]

Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = $\frac{m v^{2} }{r}$

==> 1.1 × $10^{19}$ = (9.4 × $10^{21}$ × $v^{2}$ ) ÷ 1.5 × $10^{8}$

==> 1.1 × $10^{19}$ = $v^{2}$ × 6.26× $10^{13}$

==> $v^{2}$ = 1.1 × $10^{19}$ ÷ 6.26× $10^{13}$

==> $v^{2}$ = 0.17571885 × $10^{6}$

==> v= 0.419188323 × $10^{3}$ m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

==> T= 2×3.14×1.5× $10^{8}$ ÷ 0.419188323 × $10^{3}$

==> T = 22.472× 100000 = 2247200 sec

but

86400 sec = 1 day

==> 2247200 sec= 2247200 ÷ 86400 = 26 days

3 0

1 year ago

Read 2 more answers

Lucy and her bike together have a mass of 120kg. She slows down from 4.5m/s to 3.5m/s. How much kinetic energy does she lose?

vovangra [49]

The kinetic energy of a moving object is given by
$K= \frac{1}{2}mv^2$
where m is the object's mass and v its velocity.

In our problem, the initial kinetic energy is:
$K_i = \frac{1}{2} m v_i^2 = \frac{1}{2}(120 kg) (4.5 m/s)^2=1215 J$

while the final kinetic energy is:
$K_f = \frac{1}{2}mv_f^2 = \frac{1}{2}(120 kg)(3.5 m/s)^2= 735 J$

So, the kinetic energy lost by Lucy and her bike is
$\Delta K = K_i - K_f = 1215 J - 735 J = 480 J$

7 0

1 year ago