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Rudik [331]
2 years ago
7

four students push carts filled with sports equipment across the gym. Each student pushes with the same amount of force. which c

art has the greatest change in speed
Physics
2 answers:
anastassius [24]2 years ago
6 0

The cart that has the smallest mass of sports equipment in it
has the greatest change in speed.  We know this from Newton's
second law of motion ...

                      Force = (mass) x (acceleration).

If several objects have the same force acting on them, then the one
with the smallest mass has the greatest acceleration.

Bad White [126]2 years ago
4 0
I would say the smallest car has the greatest
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What's the momentum of a 3.5-kg boulder rolling down hill at 5 m/s
ICE Princess25 [194]
P = mv 
p = 3.5 × 5 
p = 17.5 kg .m/s

Hope this helps!
6 0
2 years ago
Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
Zanzabum

(a)  3.3\cdot 10^{-6} Pa

The radiation pressure exerted by an electromagnetic wave on a surface that totally absorbs the radiation is given by

p=\frac{I}{c}

where

I is the intensity of the wave

c is the speed of light

In this problem,

I=1000 W/m^2

and substituting c=3\cdot 10^8 m/s, we find the radiation pressure

p=\frac{1000 W/m^2}{3\cdot 10^8 m/s}=3.3\cdot 10^{-6}Pa

(b) 4.4\cdot 10^{-8} m/s^2

Since we know the cross-sectional area of the laser beam:

A=6.65\cdot 10^{-29}m^2

starting from the radiation pressure found at point (a), we can calculate the force exerted on a tritium atom:

F=pa=(3.3\cdot 10^{-6}Pa)(6.65\cdot 10^{-29} m^2)=2.2\cdot 10^{-34}N

And then, since we know the mass of the atom

m=5.01\cdot 10^{-27}kg

we can find the acceleration, by using Newton's second law:

a=\frac{F}{m}=\frac{2.2\cdot 10^{-34} N}{5.01\cdot 10^{-27} kg}=4.4\cdot 10^{-8} m/s^2

6 0
2 years ago
The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the
Ray Of Light [21]

Answer:

q = 4.5 nC

Explanation:

given,

electric field of small charged object, E = 180000 N/C

distance between them, r = 1.5 cm = 0.015 m

using equation of electric field

E = \dfrac{kq}{r^2}

k = 9 x 10⁹ N.m²/C²

q is the charge of the object

q= \dfrac{Er^2}{k}

now,

q= \dfrac{180000\times 0.015^2}{9\times 10^9}

      q = 4.5 x 10⁻⁹ C

      q = 4.5 nC

the charge on the object is equal to 4.5 nC

8 0
2 years ago
Read 2 more answers
The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the
otez555 [7]

Answer:

0.647 nC

Explanation:

The force experienced by a charge due to the presence of an electric field is given by

F=qE

where

q is the charge

E is the magnitude of the electric field

In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

F=5.50\mu N = 5.50\cdot 10^{-6} N

Therefore, if the electric field magnitude is

E=8.50 kN/C = 8500 N/C

Then the charge on each antenna would be

q=\frac{F}{E}=\frac{5.50\cdot 10^{-6} N}{8500 N/C}=6.47\cdot 10^{-10} C = 0.647 nC

8 0
2 years ago
The diagram shows Ned’s movement as he left his house and traveled to different places throughout the day. Which reference point
aleksley [76]

The pet store would be the reference point because it is where he started and it will not move. Hope this helped.

6 0
2 years ago
Read 2 more answers
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