If the speed and mass of an object are doubled, which of the following is true?

An object of mass 4 kg is initially at rest on a frictionless ice rink. It suddenly explodes into three pieces (I have no idea w

Rina8888 [55]

We have two events that occurred on different axes, the most convenient is to perform the operations on the two axes, and then look for the resulting force.

Our values are,

$m_1=1kg\\m_2 = 2kg\\v_1' = 1.2m/s\\v_1''= 0.8m/s$

PAR A) For the X axis, we apply momentum conservation, which is given by,

Total momentum before = Total momentum After

We start from rest, so in X the initial speeds are 0,

$m_1v_1+m_2v_2=m_1v_1'+m_2v_2'$

$0+0 = (1*1.2)+(1)v_2'$

$V_2' = -1.2m/s$

Now we apply for the conservation of the moment, it is part of the rest, so,

$m_1v_1+m_2v_2=m_1v_1''+m_2v_2''$

$0+0=2*0.8+1*v_2''$

$v_2'' = -1.6$

To find the total speed, we simply apply pitagoras,

$V = \sqrt{v_2'^2+v_2^2''}$

$V = \sqrt{1.2^2+1.6^2}$

$V = 2m/s$

PART B) The address is given by,

$tan\theta = \frac{V_2''}{V_2'}$

$\theta = tan^{-1} \frac{V_2''}{V_2'}$

$\theta = tan^{-1} \frac{1.6}{1.2}$

$\theta = 53.31\°$ (Below -x axis)

6 0

3 years ago

Selective serotonin reuptake inhibitors began appearing in the

katrin [286]

Selective serotonin reuptake inhibitors (SSRIs) such as Prozac, Paxil and Zoloft improved the treatment of depression and other related disorders when they were introduced in the market in the 1980s. SSRIs are a type of drugs that are commonly used as antidepressants in the treatment of depressive and anxiety disorders.<span> </span>

5 0

3 years ago

Why does an astronaut in a spacecraft orbiting Earthexperience<br> a feeling of weightlessness?

slava [35]

Answer:

Astronaut in spacecraft while orbiting earth experience weightlessness because there is no gravity of earth or moon is acting on the body of an astronaut.

while on earth, we experience weight because the gravity of earth is acting on our body which is pulling us downward.

Both spacecraft and the astronauts both are in a free-fall condition.

4 0

3 years ago

Assuming that Bernoulli's equation applies, compute the volume of water ΔV that flows across the exit of the pipe in 1.00 s . In

OLEGan [10]

Answer:

discharge rate (Q) = 0.2005 m^{3} / s

Explanation:

if you read the question you would see that some requirements are missing, by using search engines, you can get the complete question as stated below:

Water flows steadily from an open tank as shown in the figure. (Figure 1) The elevation of point 1 is 10.0m , and the elevation of points 2 and 3 is 2.00 m . The cross-sectional area at point 2 is 4.80x10-2m ; at point 3, where the water is discharged, it is 1.60x10-2m. The cross-sectional area of the tank is very large compared with the cross-sectional area of the pipe. Part A Assuming that Bernoulli's equation applies, compute the volume of water DeltaV that flows across the exit of the pipe in 1.00 s . In other words, find the discharge rate \Delta V/Delta t. Express your answer numerically in cubic meters per second.

solution:

time = 1 s

elevation of point 1 (z1) = 10 m

elevation of point 2 (z2) = 2 m

elevation of point 3 (z3) = 2 m

cross section area of point 2 = 4.8 x 10^{2} m

cross section area of point 3 = 1.6 x 10^{2} m

g

acceleration due to gravity (g) = 9.8 m/s^{2}

find the discharge rate at point 3 which is the exit pipe.

discharge rate (Q) = A3 x V3

where A3 is the cross sectional area at point 3 and V3 is the velocity of the fluid and can be gotten by applying Bernoulli's equation below

$\frac{P1}{ρg} + \frac{V1^{2} }{2g} + Z1 = \frac{P3}{ρg} + \frac{V3^{2} }{2g} + Z3$

pressure at point 1 (P1) is the same as pressure at point 3 (P3), and at point 1, the velocity (V1) = 0. therefore the equation now becomes

$\frac{P1}{ρg} + Z1 = \frac{P1}{ρg} + \frac{V3^{2} }{2g} + Z3$

$Z1 = \frac{V3^{2} }{2g} + Z3$

V3 = $\sqrt{2g(Z1-Z3)}$

V3 = $\sqrt{2 x 9.8 x (10 - 3)}$

V3 = 12.53 m/s

discharge rate (Q) = A3 x V3 = 1.6 x 10^{-2} x 12.53

discharge rate (Q) = 0.2005 m^{3} / s

8 0

3 years ago

A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th

mihalych1998 [28]

We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
$a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}$
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.

4 0

2 years ago