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densk [106]
3 years ago
8

A monkey throws a banana peel from a height of 14 meters. It has a mass of 0.01 kilogram. What was the approximate potential ene

rgy of the banana peel before the monkey threw it?
Physics
1 answer:
liq [111]3 years ago
3 0
We need to use the equation for potencial energy

Ep= mgh

now we have to replace with the values we have

Ep=(14m)(9.81m/s²)(0.01kg)

Ep= 1.3734 Joules
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a block with length 1.5m width 1m height 0.5m and mass 300kg lays on the table.what is the pressure at the bottom surface of the
Nesterboy [21]

Answer:

your answer will be 320kg that would be the pressure at the bottom surface of the block

6 0
2 years ago
How could you weaken the force of gravity between cars and the Earth?<br>**I WILL MARK BRAINLEST**​
scoundrel [369]

Answer:

The answer you have selected is correct.

Explanation:

Increase radius, force of gravity decreases

6 0
3 years ago
A delivery truck leaves a warehouse and travels 3.20 km east. The truck makes a right turn and travels 2.45 km south to arrive a
boyakko [2]

Explanation:

It is given that,

Displacement of the delivery truck, d_1=3.2\ km (due east)

Then the truck moves, d_2=2.45\ km (due south)

Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :

d=\sqrt{d_1^2+d_2^2}

d=\sqrt{3.2^2+2.45^2}

d = 4.03 km

Let \theta is the direction of the truck’s displacement from the warehouse from south of east.

\theta=tan^{-1}(\dfrac{2.45}{3.2})

\theta=37.43^{\circ}

So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.

8 0
3 years ago
Free Fall: A rock is thrown directly upward from the edge of a flat roof of a building that is 56.3 meters tall. The rock misses
Slav-nsk [51]

Answer:

v₀₁= 5.525 m / s

Explanation

Freefall Formulas :

The sign of acceleration due to gravity  (g) is positive if the object is going down and negative if the object is going up.

vf= v₀+gt  

vf²=v₀²+2*g*h

h= v₀t+ (1/2)*g*t²

Where:  

h: hight in meters (m)    

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

Kinematics of the rock from the starting point with vo until it reaches its maximum height:

vf₁= v₀₁-gt₁  :vf₁ =0 to maximum height

0= v₀₁-gt₁

v₀₁ = g*t₁

t₁ =v₀₁ / g      Equation (1)

vf₁²= v₀₁²-2*g*h₁   : vf₁ =0 to maximum height

0 = v₀₁²-2*g*h₁

2*g*h₁ = v₀₁²

h₁ = (v₀₁²)/(2g)   Equation (2)

Kinematics of the rock when it falls from the maximum height until it touches the floor

h₂= v₀₂t+ (1/2)*g*t₂²  v₀₂=vf₁ =0

h₂= 0+ (1/2)*g*t₂²

h₂= (1/2)*g*t₂²   Equation (3)

Equation that relates h₁ to h₂

h₂=  h₁ + 56.3  ,  h₁ = (v₀₁²)/(2g)

h₂= (v₀₁²)/(2g) + 56.3  Equation (4)

Equation that relates t₁ to t₂

t₁ + t₂ =4 s

t₂ =4 -t₁

t₂ =4 -(v₀₁/g )

Calculation of v₀₁

We replace equation 4 and equation 5 in equation 3

(v₀₁²)/(2g) + 56.3 = (1/2)*g*(4 -(v₀₁/g ) )²

(v₀₁²)/(2g) + 56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g )+((v₀₁/g )²)

we eliminate (v₀₁²)/(2g) on both sides of the equation

56.3 = (1/2)*g* (16 - 2*4*(v₀₁/g ))

56.3 = 78.4 - 4*v₀₁

4*v₀₁ =78.4-56.3

v₀₁= (78.4-56.3) / ( 4)

v₀₁= 5.525 m / s

7 0
3 years ago
One kind of slingshot consists of a pocket that holds a pebble and is whirled on a circle of radius r. The pebble is released fr
Debora [2.8K]

Answer:

θ=142.9°

Explanation:

d=1 *r

angle ϕ= 37.1°

the line connecting pebble and target should be tangent to a circle so

cos(180-ϕ-θ)=\frac{r}{d}=\frac{1}{1}

∴ θ=180-ϕ-cos^{-1} (\frac{1}{1} )

  θ= 180-37.1-0

  θ=142.9°

 

5 0
3 years ago
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