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Advocard [28]
3 years ago
10

Question 4 of 25

Physics
1 answer:
Zinaida [17]3 years ago
5 0

Answer:

Both nuclear fission and nuclear fusion produce energy from

matter

:D

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A gardener pushes a shovel into the ground with a force of 75 N. The angle of the shovel to the ground is 80 degrees. What is th
jok3333 [9.3K]

Answer:

B

Explanation:

Resolve the 75N force into 2 components; horizontal and vertical. And remember that there is no acceleration in the downward direction, so apply Newton's second law and equate it to 0.

7 0
3 years ago
Which columns are mislabeled?
Aleks [24]

Answer:

first order date and most recent order date

Explanation:

it was switched. column 5 should be most recent order date because it's 2020 while column 6 should be first order date because it was in 2019

8 0
3 years ago
Read 2 more answers
Question 1 Gary is on the space shuttle. It takes off and lifts him to a height of 300 km above Earth's surface. a. How has Gary
balandron [24]
A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.

b) The weight of an object is given by
W=mg
where
m is the mass
g= \frac{GM}{r^2} is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.

At the Earth's surface, g=9.81 m/s^2, so Gary's weight is
W=mg=9.81 m  (1)
where m is Gary's mass.

Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is 
R=6370 km
So the distance of Gary from the Earth's center is
r=R+h=6370 km+300 km =6670 km = 6.67 \cdot 10^6 m

The Earth's mass is M=5.97 \cdot 10^{24} kg, so the gravitational acceleration is
g'=G \frac{M}{r^2}= (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )\frac{5.97 \cdot 10^{24} kg}{(6.67 \cdot 10^6 m)^2}=8.95 m/s^2

Therefore, Gary's weight at 300 km above Earth's surface is 
W' = mg' = 8.95 m (2)

If we compare (1) and (2), we find that Gary's weight has changed by
\frac{W'}{W}= \frac{8.95 m}{9.81 m}=0.91
So, Gary's weight at 300 km above Earth's surface is 91% of his weight at the surface.
6 0
3 years ago
Earth receives some of the Sun's energy in the form of heat and light.
musickatia [10]

Answer:

a) nuclear

Explanation:

8 0
3 years ago
A particle beam is made up of many protons, each with a kinetic energy of 3.25times 10-15 J. A proton has a mass of 1.673 times
ArbitrLikvidat [17]

Answer:

The magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 1.01 x 10^{-4} N/C

Explanation:

given information,

kinetic energy, KE = 3.25 x 10^{-15} J

proton's mass, m = 1.673 x 10^{-27} kg

charge, q = 1.602 x 10^{-19} C

distance, d = 2 m

to find the electric field that will stop the proton, we can use the following equation:

E = F/q

   = (KE/d) / q ,        KE = Fd --> F = KE/d

   = KE/qd

    = (3.25 x 10^{-15} J) / (1.602 x 10^{-19} C)(2 m)

    = 1.01 x 10^{-4} N/C

3 0
3 years ago
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