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gladu [14]
3 years ago
10

Match the following ERA’s Question 4

Physics
1 answer:
inna [77]3 years ago
8 0
1. Cenozoic ERA
2. Mesozoic ERA
3. Paleozoic ERA
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Could I get help on this question please . My parents won’t help me /:
vovikov84 [41]

Answer:

Tarzan will be moving at 7.4 m/s.

Explanation:

From the question given above, the following data were obtained:

Height (h) of cliff = 2.8 m

Initial velocity (u) = 0 m/s

Final velocity (v) =?

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

Finally, we shall determine how fast (i.e final velocity) Tarzan will be moving at the bottom. This can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 2.8)

v² = 0 + 54.88

v² = 54.88

Take the square root of both side

v = √54.88

v = 7.4 m/s

Therefore, Tarzan will be moving at 7.4 m/s at the bottom.

3 0
3 years ago
La resultante de dos fuerzas perpendiculares aplicadas a un mismo cuerpo es 11.18 N y el módulo de una de ellas es de 10 N. ¿Cuá
Dmitrij [34]

Answer:

English?

Explanation:

3 0
3 years ago
A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal.
Korvikt [17]

Answer:

B) 27.3 m

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x =  vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

(vfy)² = (v₀y)² - 2g(y- y₀)    Equation (2)

vfy = v₀y -gt    Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 30 m/s , at an angle  α=40.0° above the horizontal

v₀x = vx = 30*cos40° = 22.98 m/s

v₀y = 30*sin40° = 19.28 m/s

y₀ = 2m

y =  18.0 m

g = 9.8 m/s²

Calculation of the time (t) it takes for the rock to reach at  18 m above the ground

We replace data in the equation (2)

(vfy)² = (v₀y)² - 2g(y- y₀)    

(vfy)² = (19.28)² - 2(9.8)(18- 2)

(vfy)² = 371.86 - 313.6

(vfy)² = 58.26

v_{f} = \sqrt{58.26}

vfy = 7.63 m/s

We replace vfy = 7.63 m/s in the equation (2)

vfy = v₀y - gt

7.63 = 19.28 - (9.8)(t)

(9.8)(t) = 11.65

t = 11.65 / (9.8)

t = 1.19 s

Horizontal distance from where the rock was thrown to the window

We replace t = 1.19 s , in the equation (1)

x =  vx*t  

x = (22.98)* ( 1.19 )

x = 27.3 m

3 0
3 years ago
"Younger teens often brag about a lack of sleep, but that usually stops when they get older. Why do you think that is?"
svet-max [94.6K]

as we grow up we kinda experience some age staging so a teen got a lotta stuff going on they wanna be social active they be working on their popularity. so i guess thats it

3 0
3 years ago
What is the magnitude of the momentum change of two gallons of water (inertia about 7.3 kg ) as it comes to a stop in a bathtub
aliya0001 [1]

We know that the change in momentum is equals to the product of force and time that is impulse (  F \times t). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

s=ut+\frac{1}{2} gt^2

Here, u is initial velocity which is zero.

s= \frac{1}{2} gt^2 \\\\ t = \sqrt{\frac{2s}{g} }.

Thus, impulse

= F \times \sqrt{\frac{2s}{g} }

From Newton`s second law,

F =mg

Therefore, impulse

= mg \times \sqrt{\frac{2s}{g} } = m \sqrt{2gs}

Given,  m = 7.3 kg and s = 2.0 m

Substituting these values, we get

Change in momentum = impulse  

= 7.3 \ kg \sqrt{2 \times 9.8 \ m/s^2 \times 2.0 \ m } = 45 .7 \ Ns.

8 0
3 years ago
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