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Elan Coil [88]
4 years ago
8

How much time would the weight take to fall 120 m down the cliff if it was thrown downwards at 1.25 m/s?

Physics
1 answer:
Alex777 [14]4 years ago
6 0

Answer:

96 seconds

Explanation:

M/S formula

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A car starts from rest and accelerates uniformly over a time of 7 seconds for a distance of 190m. Find the the acceleration of t
Mama L [17]

Answer:

a = 7.75 [m/²]

Explanation:

To solve this problem we must use the following equation of kinematics.

x=x_{0} +v_{o} *t + (\frac{1}{2})*a*t^{2}

where:

x = final distance = 190 [m]

Xo =  initial distance = 0

Vo = initial velocity = 0 (car starts from the rest)

a = acceleration [m/s²]

t = time = 7 [s]

190 = 0 + (0*7) + 0.5*a*(7²)

190 = 0.5*49*a

a = 7.75 [m/²]

4 0
3 years ago
Technician A says that during a power balance test, the cylinder that causes the biggest RPM drop is the weak cylinder.Technicia
Anarel [89]

Answer: A I believe

Explanation:

5 0
3 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

3 0
4 years ago
What is the most likely elevation of point B?<br>a. 150 ft<br>b. 200 ft<br>c. 125 ft<br>d. 225 ft​
zepelin [54]

Answer:

D

Explanation:

erm i think its 225 ft?

7 0
4 years ago
Read 2 more answers
A rectangular tank that is 4 feet long, 2 feet wide and 15 feet deep is filled with a heavy liquid that weighs 60 pounds per cub
V125BC [204]

Answer:

a) W₁ = 54000 Lb-ft

b) W₂ = 77760 Lb-ft

c)  W₃ = 24000 Lb-ft

W₄ = 40560 Lb-ft

Step by step

W= ∫₁² ydF         1  and  2 are the levels of liquid

Where dF is the differential of weight of a thin layer

y is the height of the differential layer and

ρ*V  = F

Then

dF = ρ* A*dy*g

ρ*g  =  60 lb/ft³

A= Area of the base then

Area of the base is:

A(b) = 4*2  =  8 ft²

Now we have the liquid weighs  60 lb/ft³

Then the work is:

a)

   W₁ = ∫₀¹⁵ 8*60*y*dy     ⇒   W₁ =480* ∫₀¹⁵ y*dy

W₁ =480* y² /2 |₀¹⁵      ⇒  480/2 [ (15)² - 0 ]

W₁ = 240*225

W₁ = 54000 Lb-ft

b) The same expression, but in this case we have to pump 3 feet higher, then:

W₂ = ∫₀¹⁸ 480*y*dy     ⇒ 480*∫₀¹⁸ydy    ⇒ 480* y²/2 |₀¹⁸

W₂ =  480/2 * (18)²

W₂ =  240*324

W₂ = 77760 Lb-ft

c) To pump two-thirds f the liquid we have

2/3* 15  =  10

W₃ = 480*∫₀¹⁰ y*dy   ⇒  W₃ = 480* y²/2 |₀¹⁰

W₃ = 240*(10)²

W₃ = 24000 Lb-ft

d)

W₄ =480*∫₀¹³ y*dy

W₄ =480* y²/2 |₀¹³

W₄ = 240*(13)²

W₄ = 240*169

W₄ = 40560 Lb-ft

3 0
3 years ago
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