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3 years ago

Sceiene question please show work for brainliest :)

Physics

1 answer:

Ad libitum [116K]3 years ago

5 0

The answer to this questions would be nuclear engineer. The overall career of engineering would require physics as engineers need to know how the overall mechanics of nature affect the way their structure is built.

Hope this helped :D

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World's biggest fandom 2020

lesya692 [45]

Answer:

BTS was the biggest fandom in 2020

4 0

3 years ago

Read 2 more answers

A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic

Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light $c= 2.99793\times10^{8}\ m/s$

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

$\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}$

Put the value into the formula

$\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}$

$\Delta l=1\sqrt{1-(0.96)^2}$

$\Delta l=0.28\ m$

We need to calculate the time the stick take to move

Using formula of time

$t=\dfrac{\Delta l}{v}$

Put the value into the formula

$t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}$

$t=9.72\times10^{-10}\ sec$

$t=0.97\ ns$

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0

3 years ago

Question 2 (10 points)

kenny6666 [7]

The force required to slow the truck was -5020 N

Explanation:

First of all, we find the acceleration of the truck, which is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time

For the truck in this problem,

v = 11.5 m/s

u = 21.9 m/s

t = 2.88 s

So the acceleration is

$a=\frac{11.5-21.9}{2.88}=-3.6 m/s^2$

where the negative sign means that this is a deceleration.

Now we can find the force exerted on the truck, which is given by Newton's second law:

$F=ma$

where

m = 1390 kg is the mass of the truck

$a=-3.6 m/s^2$ is the acceleration

And substituting,

$F=(1390)(-3.6)=-5004 N$

So the closest answer among the option is -5020 N.

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

#LearnwithBrainly

8 0

3 years ago

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec

aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

$x = V*t = -8\frac{m}{s} *3s = -24m$

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

$v^2 = v_0^2 + 2a*d$

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

So, the total distance covered by this object in meters will be the sum of all the distances we found:

$x_total = 24m + 4.57m + 216.07m = 244.64m$

8 0

3 years ago

1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown

denis23 [38]

Answer:

a) 3.37 x $10^{3} kg/m^3$

b) 6.42kg/ $m^{3}$

Explanation:

a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = 50N = mg implies the mass of metal is 5kg.

Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x $10^{-4}$ $kg/m^{3}$ . So density of metal = mass of metal / volume of metal = 5 / 14 x $10^{-4}$ $kg/m^{3}$ = 3.37 x $10^{3} kg/m^3$

b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/ $m^{3}$

3 0

3 years ago