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3 years ago

Can y’all please help me with the two part question?

Physics

1 answer:

Valentin [98]3 years ago

6 0

Exactly the same situation as the baseball rolling off the desk, only with different numbers.

Distance falling from rest = (1/2) · (g) · (T²)

Janet fell from rest and landed in the water 2.1 sec later.

How high did she fall from ?

Distance = (1/2) · (9.8 m/s²) · (2.1 sec)²

Distance = (4.9 m/s²) · (4.41 s²)

<em>Distance = 21.61 meters</em> (HIGH platform. About 71 feet ! ! )

She didn't just roll off the edge of the platform.

She shot out from it horizontally, at 2.46 m/s.

She kept moving horizontally at 2.46 m/s all the way down.

How far was she from the tower when she went <em>SPLAT</em> at the bottom ?

Distance = (speed) · (time)

Distance = (2.46 m/s) · (2.1 sec)

<em>Distance = 5.17 meters</em>

<em>(That's 17 feet from the tower ! I sure hope they put the pool that far out.)</em>

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The innermost satellite of jupiter orbits the planet with a radius of 422 × 103 km and a period of 1.77 days. what is the mass o

FinnZ [79.3K]

The mass of Jupitar is obtained from the calculations as 5.8 * 10^-14 Kg.

<h3>What is the mass of Jupitar?</h3>

There are nine planets in the solar system and the sun lies at the enter of our solar system. This is the heliocentric model of the solar system.

Given that;

T^2 = GMr^3/4π

T = period

G = gravitational constant

r = radius

M = mass of Jupitar

Now;

1 day = 86400 seconds

1.77 days = 1.77 days * 86400 seconds/1 day

= 152928 seconds

Making M the subject of the formula;

M =4πT^2/Gr^3

M = 4 * 3.142 * (152928)^2/6.67 × 10^-11 * (422 × 10^9)^3

M = 2.9 * 10^11/5.0 * 10^24

M = 5.8 * 10^-14 Kg

Learn more about mass of a planet:brainly.com/question/13851553

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5 0

2 years ago

Simple Circuit and Ohm's Law Check-for-Understanding

Vanyuwa [196]

Answer:

current, only

Explanation:

current:I

voltage:U

resistance:R

formula: I=U/R

Increasing the battery cause the increasing in the voltage. Resistance does not normally change. And the current would increase.

4 0

3 years ago

Read 2 more answers

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

3 years ago

Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of

max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

$Q =\frac{KA(\delta T)}{L}$

$Q =\frac{25.87*1*20}{1}$

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0

3 years ago

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Define uniform and non uniform

cricket20 [7]

Answer:

When a body moves along a straight line with uniform speed or steady speed is called Uniform motion. When a body moves along a straight line but with variable or change in speed is called non-uniform motion.Hope this answer helps.

7 0

2 years ago

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