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3 years ago

A water tank holds water to a depth of 80 cm.

Physics

1 answer:

Vlad1618 [11]3 years ago

8 0

Answer:

8000 pa

Explanation:

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A force of 40 N accelerates a 5 block at 6 m/s ^ 2 along a horizontal surface a. What would the block's acceleration be if the s

Bas_tet [7]

Answer:

a = 8 m/s^2, Ffriction = 10 N, μk = 0.205

Explanation:

a. Force = Mass*Acceleration,

(since you didn't add the units..."5 block"....for the mass, I will assume it to be in kg, per SI units)

40 N = 5 kg*acceleration,

a = 40/5 = 8 m/s^2

b. As you know newtons second law (F=m*a) is actually in the form Fnet = m*a. Which means that if the friction force comes into play, it would be Fapplied - Ffriction = m*a.

Fapplied - Ffriction = m*a,

40 - Ffriction = 5*6,

40 - Ffriction = 30,

Ffriction = 40 - 30 = 10 N

c. The coefficient of kinetic friction is calculated by the formula "Ffriction = μk*Fnormal".

10 = μk*Fnormal (Fnormal = m*g = 5*9.8)

10 = μk*49,

μk=10/49 ≈ 0.205

7 0

3 years ago

How to get stud multipliers in lego star wars the skywalker saga?.

Anna11 [10]

You can see the Stud Multipliers right away in your Holoprojector menu under the Extras tab.

7 0

3 years ago

Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea

Setler [38]

Answer:

a) M = 4,997 10²⁰ kg , b) T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

a = (v₀ - v) / t

a = (15 - (-15)) /9.00 = 30/9

a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

F = m a

G m M / r² = m a

M = a r² / G

Let's calculate

M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

R = $R_{m}$ + h

R = 1 10⁵ + 2.00 10⁴

R = 12 10⁴ m

F = m a

G m M / R² = m a

Centripetal acceleration is

a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

d = 2π R

v = d / t

v = 2π R / T

Let's replace

G m M / R² = m (2π R / T)² / R

G M = R³ 4π² / T²

T² = 4π² R³ / G M

T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

T² = 6.82 10¹⁶ / 3.33 10¹⁰

T = √ (2,048 10⁶)

T = 1.43 10³ s

3 0

3 years ago

A gaseous system undergoes a change in temperature and volume. What is the entropy change for a particle in this system if the f

jonny [76]

Answer:

<em>Entropy Change = 0.559 Times</em>

Explanation:

Entropy change is determined by the change in the micro-states of a system. As we know that the micro-states are the same as measure of disorderness between initial and final states, that's the the amount of change in micro-states determine how much of entropy has changed in the system.

5 0

3 years ago

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

3 years ago