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2 years ago

Kindly assist on how to go about the question

Physics

1 answer:

Paul [167]2 years ago

5 0

Answer:

You need to really stop and play fortnite with me Socials 90s

Explanation:

Cuz im really good and i know i am fym

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A 7.0-kilogram cart, A, and a 3.0-kilogram cart, B, are initially held together at rest on a horizontal, frictionless surface. W

7nadin3 [17]

For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively

(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s

<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>

7 0

2 years ago

If Katie swims from one end of the pool, to the other side, and then swims back to her original spot, her average velocity is ha

Crank

Answer:

false.

Explanation:

Ok, we define average velocity as the sum of the initial and final velocity divided by two.

Remember that the velocity is a vector, so it has a direction.

Then when she goes from the 1st end to the other, the velocity is positive

When she goes back, the velocity is negative

if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:

AV = (V + (-V))/2 = 0

While the average speed is the quotient between the total distance traveled (twice the length of the pool) and the time it took to travel it.

So we already can see that the average velocity will not be equal to half of the average speed.

The statement is false

4 0

2 years ago

In the metric system, the appropriate unit for weight is the _____. gram newton newton/cm2 gram/cm3

Archy [21]

Answer:

Newton

Explanation:

The earth attracts every body towards its centre. The force with which the earth attracts any body towards its centre, is called its weight.

It is a vector quantity.

It always acts towards the centre of earth.

The SI unit of Newton.

4 0

2 years ago

A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra

skelet666 [1.2K]

Answer:

So the acceleration of the child will be $8.05m/sec^2$

Explanation:

We have given angular speed of the child $\omega =1.25rad/sec$

Radius r = 4.65 m

Angular acceleration $\alpha =0.745rad/sec^2$

We know that linear velocity is given by $v=\omega r=1.25\times 4.65=5.815m/sec$

We know that radial acceleration is given by $a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2$

Tangential acceleration is given by

$a_t=\alpha r=0.745\times 4.65=3.464m/sec^$

So total acceleration will be $a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2$

7 0

2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant