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kondor19780726 [428]
3 years ago
7

A vector always consits of

Physics
1 answer:
Kazeer [188]3 years ago
8 0

Answer:

Size and Direction

Explanation:

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Which two staments explain how a cell's parts help it get nutrients
Goryan [66]

Answer:

I think it's A and D.

6 0
2 years ago
4. As Juan is going to take a shower, the soap falls out of the soap dish on to the
LiRa [457]

The coefficient of friction between the soap and the floor is 0.081

If Juan steps on the soap with a force of 493 N, this is her weight, W. This weight also equals the normal reaction on the floor, N.

We know that frictional force F = μN where μ = coefficient of friction between soap and floor.

So, μ = F/N

Since F = 40 N and N = W = 493 N,

μ = F/N

μ = 40 N/493 N

μ = 0.081

So, the coefficient of friction between the soap and the floor is 0.081

Learn more about coefficient of friction here:

brainly.com/question/13923375

5 0
3 years ago
Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of 60 kV.
ivolga24 [154]

Answer: 2.068*10^{-14}m

Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.

workdone= qV

energy = hc/λ

q=magnitude of an electronic charge= 1.602*10^{-16}

h= planck constant = 6.626*10^{-34}

c= speed of light =2.998* 10^{8}

v= potential difference= 6*10^{4}

λ= wavelength=unknown

by making λ subject of formulae we have that

λ= \frac{hc}{qv}

λ = 6.626*10^{-34} * 2.998* 10^{8} / 1.602*10^{-16} * 6*10^{4}

λ = \frac{19.878*10^{-26} }{9.612*10^{-12} }

by doing the necessary calculations, we have that

λ = 2.068*10^{-14}m

8 0
3 years ago
A pendulum
RSB [31]

Explanation:

The time period rotation of the bob is t and the tension in the thread is T. ... A simple pendulum of length. ... Balancing the forces in Horizontal and vertical direction: ... Circular motion 2.

1 answer

8 0
3 years ago
Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
2 years ago
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