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4 years ago

10

GETS BRAINILIST!!!!Which of the following will increase the gravitational force exerted by one object on another? Increasing the

temperature of the objects Decreasing the temperature of the objects Decreasing the distance between the objects Increasing the distance between the objects

1 answer:

jeka944 years ago

5 0

Answer:

C. Decreasing the distance between the objects.

Explanation:

Gravitational force depends on mass and distance. The farther the objects are apart, the less gravitational force/pull they have.

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A bike race against the clock takes place on a straight road. Yan drives at 37 km / h and he starts the course 30s before Christ

joja [24]

Given data:

Yan speed;

$u_1=37\text{ km/h}$

Christopher speed;

$u_2=38.9\text{ km/h}$

Christophe starts 30 s later than Yan. Therefore, Christophe takes 30 s less than Yan to reach the same distance.

Part (A)

The distance is given as,

$d=ut$

Let both Yan and Christophe meet at d distance from the start position. Therefore,

$u_1t=u_2(t-30)$

Substituting all known values,

$\begin{gathered} (37\text{ km/h})t=(38.9\text{ km/h})\times(t-30) \\ \frac{(37\text{ km/h})}{(38.9\text{ km/h})}=\frac{(t-30)}{t} \\ 0.95=1-\frac{30}{t} \\ \frac{30}{t}=1-0.95 \\ \frac{30}{t}=0.05 \\ t=\frac{30}{0.05} \\ t=600\text{ s} \end{gathered}$

Therefore, 600 s after Yan's departure Christophe will join him.

Part (B)

The distance is given as,

$d=u_1t$

Substituting all known values,

$\begin{gathered} d=(37\text{ km/h})\times(600\text{ s}) \\ =(37\text{ km/h})\times(600\text{ s})\times(\frac{1\text{ hr}}{3600\text{ s}}) \\ \approx6.17\text{ km} \end{gathered}$

Therefore, Christophe joins Yan after 6.17 km from the start.

3 0

1 year ago

How do you do this problem?

kvasek [131]

Explanation:

First, find the velocity of the projectile needed to reach a height h when fired straight up.

Given:

Δy = h

v = 0

a = -g

Find: v₀

v² = v₀² + 2aΔy

(0)² = v₀² + 2(-g)(h)

v₀ = √(2gh)

Now find the height reached if the projectile is launched at a 45° angle.

Given:

v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)

v = 0

a = -g

Find: Δy

v² = v₀² + 2aΔy

(0)² = √(gh)² + 2(-g)Δy

2gΔy = gh

Δy = h/2

5 0

3 years ago

A simple series circuit consists of a 130 ? resistor, a 30.0V battery, a switch, and a 2.10 pF parallel-plate capacitor (initial

V125BC [204]

Answer with Explanation:

We are given that

Resistance,R=130 ohm

Potential difference, V=30 V

Capacitor,C=2.1pF= $2.1\times 10^{-12} F$

$1pF=10^{-12} F$

$d=5.0 mm=5\times 10^{-3} m$

$1mm=10^{-3} m$

a.Maximum flux

$\phi=EA=\frac{V}{d}\times \frac{Cd}\epsilon_0}=\frac{CV}{\epsilon_0}$

$\epsilon_0=8.85\times 10^{-12}$

$\phi=\frac{2.1\times 10^{-12}\times 30}{8.85\times 10^{-12}}=7.12Vm$

b.Maximum displacement current, $I=\frac{V}{R}=\frac{30}{130}=0.23 A$

c.We have to find electric flux at t=0.5 ns

$t=0.5ns=0.5\times 10^{-9} s$

$1ns=10^{-9}s$

$q=CV(1-e^{-\frac{t}{RC}})$

$q=30\times 2.1\times 10^{-12}(1-e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-9}})$

$q=52.9\times 10^{-12} C$

$\phi=\frac{q}{\epsilon_0}=\frac{52.9\times 10^{-12}}{8.85\times 10^{-12}}=5.98Vm$

d.Displacement current at t=0.5ns

$I=(\frac{V}{R})e^{-\frac{t}{RC}}=\frac{30}{130}e^{-\frac{0.5\times 10^{-9}}{130\times 2.1\times 10^{-12}}}$

$I=0.037 A$

5 0

3 years ago

The image shows a pendulum that is released from rest at point A. Which energy transformation is correct? From A to C, kinetic e

irakobra [83]

Answer:

From C to D, kinetic energy is transformed into gravitational potential energy.

i think

8 0

3 years ago

Read 2 more answers

Part 4! Anything helps!!! Please and thanks

lisov135 [29]

1) 25 because I did 5*5
2) It would be the third option because a longer wire has a higher resistance and a thinner wire causes electrons to travel at a slower rate

Not sure about the last one, sorry

4 0

4 years ago