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3 years ago

Prečo je omnoho ľahšie udržať sa na hladine vody v mori ako v rieke

Physics

1 answer:

Vika [28.1K]3 years ago

6 0

Answer:

please write in english language so that we can help you

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A brick is dropped from a high scaffold. a. What is its velocity after 4.0s ?

Ilia_Sergeevich [38]

Answer:

A: 1.962

B: 3.924

Explanation:

g = G *M /R^2

g = 9.807*M/R^2 the gravitational constant of ground level on earth is about 9.807

g = 9.807*5lbs/R^2 the average brick is about 5 pounds.

g = 9.807*5*10^2. I'm assuming the height is around ten feet to help you out.

with these numbers plugged in you get an acceleration of 0.4905 a final velocity after 4 seconds 1.962. It's height fallen after 4 seconds is 3.924.

( M = whatever the brick weighs it's not specified in the question)

(R = the distance from the ground or how high the scaffold is)

(hopefully you can just plug your numbers in there hope this helps)

6 0

3 years ago

A plate carries a charge of 3.8 UC, while a rod carries a charge of 1.9 C. How many electrons must be transferred from the plate

frosja888 [35]

Answer:

$N_{electrons}=Q_{transfered}/q_{electron}=5.94*10^{18}electrons$

Explanation:

The total charge is distributed over the two objects:

$Q_{total}/2=(3.8*10^{-6}C+1.9C)/2=0.9500019C\\$

The plate and the rod must have $Q_{total}/2\\$ . So the charge transferred from the plate to the rod is:

$Q_{transfered}=3.8*10^{-6}C-Q_{total}/2=3.8*10^{-6}C-0.9500019C=-0.9499981C\\$

Number of electrons:

$N_{electrons}=Q_{transfered}/q_{electron}=-0.9499981C/(-1.6*10^{-19}C)=5.94*10^{18}electrons$

4 0

2 years ago

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

puteri [66]

Answer:

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

Explanation:

Given:

Range of speed during which constant power is supplied to the wheels by the car is $0\ mph\ to\ 70\ mph$ .

Initial velocity of the car, $v_i=0\ mph$

final velocity of the car during the test, $v_f=32\ mph=14.3052\ m.s^{-1}$

Time taken to accelerate form zero to 32 mph at full power, $t=1.2\ s$

initial velocity of the car, $u_i=0\ mph$

final desired velocity of the car, $u_f=64\ mph=28.6105\ m.s^{-1}$

Now the acceleration of the car:

$a=\frac{v_f-v_i}{t}$

$a=\frac{14.3052-0}{1.2}$

$a=11.921\ m.s^{-1}$

Now using the equation of motion:

$u_f=u_i+a.t$

$64=0+11.921\times t$

$t=5.3687\ s$ is the time taken by the car to accelerate the desired range of the speed from zero at full power.

8 0

2 years ago

After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position

KiRa [710]

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

$W_p = KE_f - KE_i$

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

$KE_f = \frac{1}{2}mv_1^2$

now work done is given as

$W_p = \frac{1}{2}mv_1^2 - 0$

so we can say

$W_p = \frac{1}{2}mv_1^2$

so above is the work done on the box to slide it from x1 to x2

3 0

3 years ago

PLEASE HELP!!!!! (20 points!)

valentina_108 [34]

Step 2
step 5
step 4
step 6
step 1
step 3

thank me later

3 0

3 years ago

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