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Natalija [7]
3 years ago
6

43. A mother pushes her 120-newton child, who is

Physics
1 answer:
Pepsi [2]3 years ago
8 0

5.0 N.s ( 5.0 newton second) is the magnitude of the impulse imparted  to the child by the mother

Option A

<u>Explanation:</u>

An impulse is a variation in the object's momentum when the object is subjected to force for a time period. Therefore, you can use the pulse to calculate the change in pulse or use the pulse to calculate the average impact force. The equation for impulse is given by,

        J= Force \times time

Given data:

F = 10 N

t = 0.50 s

We need to find the magnitude of the impulse. By applying the given values to the above equation, we get

             J = 10 \times 0.5=5.0\ \mathrm{N} .\mathrm{s}

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A fire woman dropped a person onto the safety net. Right
dolphi86 [110]

Answer:

m = 28.7[kg]

Explanation:

To solve this problem we must use the definition of kinetic energy, which can be calculated by means of the following equation.

E_{k}=\frac{1}{2}*m*v^{2}\\

where:

Ek = kinetic energy = 1800 [J]

m = mass [kg]

v = 11.2 [m/s]

1800=\frac{1}{2}*m*(11.2)^{2}\\m = 28.7[kg]

7 0
2 years ago
b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v
velikii [3]

Answer:

\theta = n\pi/2, {\rm where~n~is~an~integer.}

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k

Similarly,

\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)

So,

\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)

At time t = 0, this equation becomes

0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}

7 0
2 years ago
Which equation can be used to calculate the normal force on an object if you know the speed of the object, the coefficient of ki
tino4ka555 [31]

Answer:

D

Explanation:

The friction force is the weight force times the coefficient of friction. So diving by the coefficient gives you the weight force which is equivalent to the normal force.

3 0
1 year ago
A 200kg bucket of cement<br>​
ozzi

Answer:

Yes. A 200 kg bucket of cement = About 440.925 pounds of cements.

Explanation:

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