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Digiron [165]
3 years ago
11

Two arrows are fired horizontally with the same speed of

Physics
1 answer:
Fed [463]3 years ago
6 0

Answer:

Explanation:

Given

mass of each arrow=0.1 kg

velocity of arrow=30 m/s

One arrow is fired u=due to east and another towards south

Momentum of first arrow

P_1=0.1\left ( 30\hat{i}\right )=3\hat{i}

P_2=0.1\left ( -30\hat{j}\right )=-3\hat{j}

Total momentum P

P=P_1+P_2

P=3\hat{i}-3\hat{j}

magnitude |P|=\sqrt{3^2+3^2}=3\sqrt{2}

direction

tan\theta =\frac{-3}{3}

\theta =45^{\circ} clockwise w.r.t to east

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A remote Internet Web page may sometimes reach your computer by going through a geostationary satellite orbiting approximately 3
soldier1979 [14.2K]

Answer:

0.24 seconds

Explanation:

The signal will reach the computer at light speed = 3\times 10^8\ m/s

Distance between the satellite and the computer = 3.6\times 10^7m

Since, the bodies have no acceleration relative to each other we use the following formula

Time = Distance / Time

\text{Time}=\frac{3.6\times 10^7}{3\times 10^8}\\\Rightarrow \text{Time}=0.12\ s

One way delay would be at least 0.12 seconds.

But when you click on a link to open a website this signal has to be first sent to the satellite then the satellite would have to send the required signal back to the computer.

Hence, the minimum time delay would be 0.12+0.12 = 0.24 seconds

3 0
3 years ago
In a "worst-case" design scenario, a 2000-{\rm kg} elevator with broken cables is falling at 4.00{\rm m/s} when it first contact
OLga [1]

Answer:

v=3.73m/s,   a=3.35m/s^{2}

Explanation:

In order to solve this problem, we must first draw a diagram that will represent the situation (See attached picture).

So there are two ways in which we can solve this proble. One by using integrls and the other by using energy analysis. I'll do it by analyzing the energy of the system.

So first, we ned to know what the constant of the spring is, so we can find it by analyzing the energy on points A and C. So we get the following:

K_{A}+U_{Ag}=K_{C}+U_{Cs}+U_{Cg}+W_{Cf}

where K represents kinetic energy, U represents potential energy and W represents work.

We know that the kinetic energy in C will be zero because its speed is supposed to be zero. We also know the potential energy due to the gravity in C is also zero because we are at a height of 0m. So we can simplify the equation to get:

K_{A}+U_{Ag}=U_{Cs}+W_{Cf}

so now we can use the respective formulas for kinetic energy and potential energy, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}ky_{C}+fy_{C}

so we can now solve this for k, which is the constant of the spring.

k=\frac{mV_{A}^{2}+mgh_{A}-fy_C}{y_{C}^{2}}

and now we can substitute the respective data:

k=\frac{(2000kg)(4m/s)^{2}+(2000kg)(9.8m/s^{2})(2m)-(17000N)(2m)}{(2m)^{2}}

Which yields:

k= 9300N/m

Once we got the constant of the spring, we can now find the speed of the elevator at point B, which is one meter below the contact point, so we do the same analysis of energies, like this:

K_{A}+U_{Ag}=K_{B}+U_{Bs}+U_{Bg}+W_{Bf}

In this case ther are no zero values to eliminate so we work with all the types of energies involved in the equation, so we get:

\frac{1}{2}mV_{A}^{2}+mgh_{A}=\frac{1}{2}mV_{B}^{2}+\frac{1}{2}ky_{B}^{2}+mgh_{B}+fy_{B}

and we can solve for the Velocity in B, so we get:

V_{B}=\sqrt{\frac{mV_{A}^{2}-ky_{B}^{2}+2mgh_{B}-2fy_{B}}{m}}

we can now input all the data directly, so we get:

V_{B}=\sqrt{\frac{(2000kg)(4m/s)^{2}-(9300N/m)(1m)^{2}+2(2000kg)(9.8m/s^{2})(1m)-2(17000N)(1m)}{(2000kg)}}

which yields:

V_{B}=3.73m/s

which is the first answer.

Now, for the second answer we need to find the acceleration of the elevator when it reaches point B, for which we need to build a free body diagram (also included in the attached picture) and to a summation of forces:

\sum F=ma

so we get:

F_{s}+f-W=ma

Supposing the acceleration is positive when it points upwards.

So we can solve that for the acceleration, so we get:

a=\frac{F_{s}+f-W}{m}

or

a=\frac{ky_{B}+f-W}{m}

and now we substitute the data we know:

a=\frac{(9300N/m)(1m)+(17000N)-(2000kg)(9.8m/s^{2})}{2000kg}

which yields:

3.35m/s^{2}

8 0
4 years ago
find the coefficient of kinetic friction for a 10 kg box being dragged steadily across the surface with a force of 2.0 Newtons​
Mazyrski [523]

Answer:

0.02

Explanation:

Force, F=\mu_k N where N is normal reaction and coefficient of kinetic friction is \mu_k. Also, N is equivalent to mg ie N=mg where m is mass of an object and g is acceleration due to gravity. Making \muthe subject of the formula then

\mu_k=\frac {F}{mg}

Substituting F with 2 N, m with 10 kg and g with 9.81 then

\mu_k=\frac {2}{10\times 9.81}=0.0203873598369\approx 0.02

5 0
4 years ago
a student determines the circumference of a golf ball. a man pulls a sledge of mass 25 kg across level ground with a horizontal
Travka [436]

Explanation:

Fnet = Ft - Ff = 60N - 20N = 40N

a = Fnet / m = 40N / 25kg = 1.6m/s².

4 0
3 years ago
Read 2 more answers
Three adjacent keys on a piano (F, F-sharp, and G) are struck simultaneously, producing frequencies of 349, 370, and 392 Hz. Wha
PSYCHO15rus [73]

Answer:

21 Hz, 43 Hz and 22 Hz

Explanation:

The computation of the beat frequencies that are generated by this discordant combination is as follows:

As we know that

beat frequencies = |f_1  - f_2|

So

For the first one

= |349 Hz - 370 Hz|

= 21 Hz

For the second one

= |349 Hz - 392 Hz|

= 43 Hz

And, for the third one

= |370 Hz - 392 Hz|

= 22 Hz

5 0
3 years ago
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