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3 years ago

A student drops a rock from a bridge to the water 12m below. How many seconds does it take the rock to hit the water?

1 answer:

7 0

R = $\frac{g t^{2} }{2}$
r - displacement or height
t - time taken
g = 10 m/s-2
t = square root from 2r/g = $\sqrt{ \frac{2 x 12}{10} }$ = 1.5 seconds

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An astronaut is in equilibrium when he is positioned 140 km from the center of asteroid C and 581 km from the center of asteroid

serg [7]

Answer:B

Explanation:

Given

Distance of astronaut From asteroid x is $r_x=140 km$

Distance of astronaut From asteroid Y is $r_y=581 km$

Suppose M,M_x,M_y be the masses of Astronaut , asteroid X and Y

If the astronaut is in equilibrium then net gravitational force on it is zero

$F_x=F_y$

$\frac{GMM_x}{r_x^2}=\frac{GMM_y}{r_y^2}$

cancel out the common terms we get

$\frac{M_x}{r_x^2}=\frac{M_y}{r_y^2}$

$\frac{M_x}{M_y}=(\frac{r_x}{r_y})^2$

$\frac{M_x}{M_y}=(\frac{140}{581})^2$

$\frac{M_x}{M_y}=0.05806\approx 0.0581$

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3 years ago

Two insulated current-carrying wires (wire 1 and wire 2) are bound together with wire ties to form a two-wire unit. The wires ar

AysviL [449]

Answer:

I'm not sure, My best friend knows, but I don't...

Explanation:

I'm sorry :)

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2 years ago

A glass falls from a table top and smashes on the floor 0.6 seconds later. How high is the table?

soldi70 [24.7K]

Answer:

1.8 m

Explanation:

Given: Glass falls from a table, smashes 0.6 seconds later

To find: How high a table is

Formula: Vv=gt, dv=1/2gt^2, t=2d/g

Solution: A table's height is measured from the top of the edge down to the floor. The tables are shown both have a height of 30 inches, which is common for many tables.

Data

t = 0.6s

g = 9.81 m/s²

d = ?

Equation

d = $\frac{1}{2}$ $gt$ ²

Math & Units

d = 4.905 (0.6²)
d = 442.676

Hence the table is 1.8 m high

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3 years ago

A recipe for a sweet tea calls for 3 cups of sugar when you place a sugar into the tea it doesn't dissolve. How could you use di

barxatty [35]

Answer: You could dissolve it by heating it back up, then just cooling it down again.

Hope that helps!

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2 years ago

Read 2 more answers

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago