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anastassius [24]

2 years ago

5

A fish is swimming around the 720-meter perimeter of her pond. If she swims 10 laps in 120 minutes, what is her average speed in

m/min

2 answers:

natulia [17]2 years ago

6 0

Her average speed is 60 m/min

zavuch27 [327]2 years ago

5 0

Answer:

60

Explanation:

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Suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a

EleoNora [17]

Newton's second law ...Force = momentum change/time.momentum change = Forcextme.also, F=ma -> a=F/m - the more familiar form of Newton's second law
using one of the kinematic equations for m ... V=u+at; u=0; a=F/m -> V=(F/m)xt.-> t=mV/F using one of the kinematic equations for 2m ... V=u+at; u=0; a=F/2m -> V=(F/2m)xt. -> t=2mV/F (twice as long, maybe ?)
I think I've made a mistake somewhere below, but I think that the principle is right ...using one of the kinematic equations for m ... s=ut + (1/2)at^2); s=d;u=0;a=F/m; t=1; -> d=(1/2)(F/m)=F/2musing one of the kinematic equations for 2m ... s=ut + (1/2)at^2); s=d;u=0;a=F/2m; t=1; -> d=(1/2)(F/2m)=F/4m (half as far ????? WHAT ???)

3 0

2 years ago

Read 2 more answers

Suppose Earth's mass increased but Earth's diame-

navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

$F = G \frac{m1m2}{r^2}$

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0

2 years ago

A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i

enyata [817]

Answer:

Emitted power will be equal to $7.85\times 10^{-5}watt$

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing $I=10^{-12}W/m^2$

We have to find the emitted power

Emitted power is equal to $P=I\times A$

$P=I\times 4\pi R^2$

$P=10^{-12}\times 4\times 3.14\times 2500^2=7.85\times 10^{-5}watt$

So emitted power will be equal to $7.85\times 10^{-5}watt$

4 0

2 years ago

What keeps an inflated balloon from falling down if you rub it against your hair and place it against a wall? 1. rubbing distort

Keith_Richards [23]

2 because when you are doing this it causes friction Which then cause the balloon to stick
~dany-ley

3 0

2 years ago

A vertical cylinder is divided into two parts by a movable piston of mass m. The piston and cylinder system is well insulated (t

Mekhanik [1.2K]

Answer:

Final temperature will be 438.076 K

Explanation:

We have given temperature $T_1=323K$

Volume $V_1=V\ and\ V_2=\frac{V}{2}$

As there is no heat transfer so this is an adiabatic process

For and adiabatic process $TV^{\gamma -1}=constant$

Here $\gamma =1.4$

So $T_1V_1^{\gamma -1}=T_2V_2^{\gamma -1}$

$T_2=\left ( \frac{V_1}{V_2} \right )^{\gamma -1}\times T_1$

$T_2=\left ( \frac{V}{\frac{V}{2}} \right )^{1.4 -1}\times 332=2^{0.4}\times 332=438.076K$

4 0

2 years ago