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2 years ago

Use one of the following formulas to solve distance:

Physics

1 answer:

zhuklara [117]2 years ago

4 0

Answer:

72m

Explanation:

Given parameters:

Speed of the Burrito = 4m/s

Time = 18s

Unknown:

Distance covered by the flying Burrito = ?

Solution:

Speed is the rate of distance covered per unit of time.

Speed = $\frac{distance}{time}$

Distance = Speed x time

So;

Distance = 18 x 4 = 72m

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When the distance between two masses is doubled, the gravitational attraction between

mrs_skeptik [129]

The new gravitational attraction will be 1/4 as much

Explanation:

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, the original force between the two objects is F, when they are separated by a distance r.

Later, the distance between the two objects is doubled, so the new distance is

$r'=2r$

Therefore, the new force will be

$F'=G\frac{m_1 m_2}{(2r)^2}=\frac{1}{4}(\frac{Gm_1 m_2}{r^2})=\frac{F}{4}$

Therefore, the new force will be one-fourth as much.

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

5 0

2 years ago

A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp

Kisachek [45]

Answer:

The force is $F= 46.25kN$

Explanation:

The diagram for this question is shown on the first uploaded image

At Equilibrium the summation of the of force on the vertical axis is zero

i.e $\sum F_y =0$

=> $F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)$

$v_2$ is the is the speed of water at the nozzle which can be mathematically evaluated as

$v_2 = \frac{R}{A_n}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(12*\frac{1m}{100} )^2$ for $A_n$

$v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }$

$= 44.23 m/s$

$v_1$ is the is the speed of water at the pipe which can be mathematically evaluated as

$v_1 = \frac{R}{A_p}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(30*\frac{1m}{100} )^2$ for $A_p$

$v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }$

$= 7.07 m/s$

$\rho$ is he density of water with value $\rho =1000 kg /m^3$

Substituting values into the equation above

$F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)$

$= 21.99kN$

At Equilibrium the summation of the of force on the horizontal axis is zero

i.e $\sum F_x =0$

=> $F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)$

Since The speed at both A and B nozzle are the same then $v_2$ remains the same

Substituting values

$F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)$

=> $F_x = 40.69kN$

Hence the force acting on the flange bolts required to hold the nozzle in place is

$F = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{40.69 ^2 + 21.99^2}$

$F= 46.25kN$

6 0

2 years ago

Read 2 more answers

A Styrofoam ball has just been shot straight up. Air resistance is not negligible (free body diagram)

timofeeve [1]

In a free body diagram for an object projected upwards;

the acceleration due to gravity on the object is always directed downwards.

the velocity of the object is always in the direction of the object's motion.

An object projected upwards is subjected to influence of acceleration due to gravity.

As the object accelerates upwards, its velocity decreases until the object reaches maximum height where its velocity becomes zero and as the object descends its velocity increases, which eventually becomes maximum before the object hits the ground.

To construct a free body diagram for this motion, we consider the following;

the acceleration due to gravity on the object is always directed downwards

the velocity of the object is always in the direction of the object's motion.

<u>For instance:</u>

upward motion for velocity ↑ downward motion for velocity ↓

↑ ↓

acceleration due to gravity ↓

↓

Learn more here: brainly.com/question/13235430

5 0

2 years ago

One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.

Zepler [3.9K]

Answer:

The answer is " $\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}$ "

Explanation:

Given:

$\to v=1\ liter= 10^{-3} \ m^3\\\\\to w= 9.6 \ N\\$

calculation:

$Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978$

4 0

2 years ago

5. __________ is the idea that all people must follow the law and that the law is applied to all equally even people in the gove

padilas [110]

The answer is Rule of the law

5 0

2 years ago