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Rufina [12.5K]
3 years ago
14

Which of the following is a vector quantity?

Physics
2 answers:
mestny [16]3 years ago
5 0
A is the answer  of course
olga55 [171]3 years ago
5 0

D. 55mph to the south

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What are the characteristics of high energy waves?
sladkih [1.3K]
<span>The relationship between wavelength, frequency and energy of Electromagnetic Radiation is given by E = hf = hc/lamba -------(1) So from (1) there's a linear relationship between E and f. The higher the frequency, f, the higher the energy E. Also from (1) it is obvious that the lower the wavelength, lambda, the higher the energy, E. This means the answer is D.</span>
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3 years ago
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You pull a block of mass m across across a frictionless table with a constant force. you also pull with an equal constant force
KiRa [710]
For any mass m:

a = F/m
v = √2*F/m*s = √2F/sm = k/√m
Momentum = mv = k√m
Energy = 1/ mv² = 1/2 m.k²/m = 1/2k²

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4 0
3 years ago
Calculate the speed of a bike rider who accelerates (from rest) for 5 seconds down a hill at an acceleration of 8
olya-2409 [2.1K]

Answer:

speed=abs(v)=40ms^-1

Explanation:

acceleration, a = (v-u)/t

since initial velocity u=0 (at rest) and a=8ms^-2,

8=v/5

hence after 5 seconds, v=40ms^-1

6 0
3 years ago
A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is
OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

U = \frac{GMm}{r}

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}

U_2= - 5.1180*10^9J

Point B )

We now use the average radius distance from the earth.

U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}

U_2= -5.8925*10^9J

Then,

\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)

\Delta U = 774.5*10^6

By the law of conservation of energy we know that,

\Delta U = \frac{1}{2}mv^2

clearing v,

v= \sqrt{2 \Delta U/m}

v= \sqrt{2*774.5*10^6 /94.2}

v= 4055.08m/s

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0
3 years ago
at stop light the truck traviling at 15m/s passes the carb as it starts from rest the truck travels at a constant relatively ,th
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What's the velocity?

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