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mars1129 [50]
3 years ago
5

A family is skating at an ice rink. The 58.2 kg mother is holding the

Physics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

When I got this question I had to draw it out so if you have to do that, draw 3 stick figures holding hands, one representing the mother, father, and daughter. Then you write their weights on top of them and then draw an arrow pointing from the father to the mother.

Explanation:

use this formula :

a_{y} = \frac{Fdadshandy}{msys}

then you fill it in :

a_{y} = \frac{100N}{35.5kg+58.2kg}

a_{y} = \frac{100N}{93.7kg}

a_{y} = 1.0672 m/s^{2}

then you multiply that with the daughters weight :

T_{2} x= m_{2} a_{y}

T_{2} x = 35.5kg (1.0672 m/s^{2})

T_{2} x = 37.89N

and that's the answer :) : 37.89N

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What is the formula to find initial velocity?​
Inga [223]

Answer:

s = ut + 1/2at^2

Explanation:

s= distance

u= initial velocity

a= acceleration

7 0
2 years ago
A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut
blsea [12.9K]

Answer:

1.E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2.E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

\rho (r) = \alpha (1-\frac{r}{R})

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

\int\, da = 2\pi r h

where ‘h’ is the length of the imaginary Gaussian surface.

Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3. At the boundary where r = R:

E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}

As can be seen from above, two E-field values are equal as predicted.

4 0
3 years ago
A rising pendulum bob gains _____ energy.
puteri [66]

Answer:

B. Kinetic energy.............

4 0
2 years ago
Read 2 more answers
A dolphin swims 56 meters in 8 seconds and a walrus swims 30 meters in 6 seconds. Which is one has the faster speed, the dophin
Natali [406]

Answer:

dolphin= 7 meters/1 second      walrus= 5 meters/1 second

Explanation:

56 divided by 8 is 7

30 diviided by 6 is 5

5 0
3 years ago
A galaxy spectrum has a redshift of 70,000 km per second. If the Hubble constant is 70 km per second per Mpc (megaparsec), how f
gogolik [260]

Using the Hubble law v = H₀d where v = recessional speed = 70,000 km per second H₀ = hubble constant = 70 km/s/Mpc and d = distance of galaxy.

Making d subject of the formula, we have

d = v/H₀

Substituting the values of the variables into the equation, we have

d = v/H₀

d = 70000 km/s/70 km/s/Mpc

d = 1000 Mpc

So, the galaxy is 1000 Mpc away from us.

Learn more about hubble law here:

brainly.com/question/18484687

8 0
2 years ago
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