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2 years ago

14

Please help I'm stuck on this question

1 answer:

Afina-wow [57]2 years ago

8 0

Answer:

increase

decrease

Explanation:

using formula

Vt=mg/6πηr

so if m increases V increases

r is the denominator so if r increases V decreases

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1. The farthest star listed in the chart is? Arcturus, Capella, Aldebaran, Pollux, Regulus, Castor

Pachacha [2.7K]

Answer:

Regulus

Explanation:

Which is 77.63 light years away

8 0

2 years ago

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .

zimovet [89]

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

$F=m\times a$

$F=m\times (\dfrac{v-u}{t})$

$F=0.003\times (\dfrac{30}{0.06})$

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

6 0

2 years ago

The strength of an electromagnet CANNOT be increased by?

zmey [24]

Answer:

reversing the current

Explanation:

4 0

2 years ago

Any child is pushing a shopping cart at a speed of 1.5 m/s.how long will it take this child to push the cart down the aisle with

NARA [144]

1.5 m/s is the velocity. 9.3 m is the length of aisle, over which Distance will be covered. Time is demanded in which the child will move the cart over the aisle with 1.5 m/s. v=S/t and, t=S/v Put values, t=9.3/1.5=6.2 s

7 0

2 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago