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Firdavs [7]
2 years ago
6

HELP ASAP PLEASE!!!

Physics
2 answers:
Montano1993 [528]2 years ago
7 0

Answer:

The answer is A, B, C, D

Explanation:

This is because gravity is the weakest force of the four fundamental forces, so it automatically cancels letter E

Vesnalui [34]2 years ago
6 0
It’s A,B,C,D
hope this helps
You might be interested in
Which statement corresponds to emission spectra?
monitta
B
i just took the test






7 0
3 years ago
A ball is shot from the ground into the air. At a height of 8.8 m, the velocity is observed to be
Mariulka [41]

Answer:

h = 10.4 m

R = 22.48 m

v= 16,2 m/s , α = 61.7°, below the horizontal

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

Explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t  :

Equations of the uniformly accelerated rectilinear motion of upward   (vertical ).

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m  

y: vertical position in meters (m)  

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Known data

y= 8.8 m

v = ( (7.7)i + (5.7)j  ) m/s : vx= 7.7 m/s , vy= 5.7 m/s

g = 9.8 m/s²

Calculation of the  initial  vertical velocity ( v₀y)

We apply Equation (3) with the known data

(vfy)² = (v₀y)² -2*g*y

(5.7)² = (v₀y)²- (2)*(9.8)*(8.8)

(5.7)²+ 172.48 =  (v₀y)²

v_{oy} = \sqrt{(5.7)^{2}+ 172.48 }

v₀y = 14.3 m/s

Calculation of the maximum height  the ball rise (h)

In the maximum height vfy=0

We apply the Equation (3) :

(vfy)² = (v₀y)² -2*g*y

0 = (14.3)² - 2*98*h

h = (14.3)² / 19.6

h = 10.4 m

Calculation of the time it takes for the ball to the maximum height

We apply the Equation (4) :

vfy = v₀y -gt

0 = v₀y -gt

gt = v₀y

t = v₀y/g

t = 14.3/9.8

t= 1.46 s

Flight time = 2t = 2.92 s

Total horizontal distance traveled by the ball  (R)

We replace data in the equation (1)

x =vx*t    vx= 7.7 m/s , t =2.92 s  (Flight time)

R = (7.7)* (2.92) = 22.48 m

Velocity of the ball (magnitude (v) and direction (α)) the instant before it hits the ground

vx = 7.7 m/s

vy = v₀y -gt = 14.3 - 9.8* (2.92) = -14.3 m/s

v= \sqrt{v_{x}^{2}+v_{y}^{2}  }

v= \sqrt{(7.7)^{2}+ (-14.3)^{2}  }

v= 16,2 m/s

\alpha = tan^{-1} (\frac{v_{y} }{v_{x} })

\alpha = tan^{-1} (\frac{-14.3 }{7.7 })

α = -61.7°

α = 61.7°, below the horizontal

i- j components of the v

v = (7.7)i + (-14.3)j in meters per second (i horizontal, j downward)

5 0
2 years ago
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 3.9 m/s. The car is a distanc
sammy [17]
If it takes t seconds to reach the car, then the distance d is 3.9t.

The bear's distance from the tourist's starting point is
6t-23

For maximum d, we set the equations equal to each other:
3.9t=6t-23
\Rightarrow -2.1t=-23
\Rightarrow t=\frac{23}{2.1}
so the distance is
d=3.9(\frac{23}{2.1})\approx42.741\ m
6 0
3 years ago
Help me calculate the kinetic energy (just the middle column) ASAP! SHOW WORK! ON PAPER
lukranit [14]

Answer:

Explanation:

I can tell you what the answers for the middle column are, but if you don't know how to solve total energy problems, they won't make any sense to you at all.

First row, KE = 0

Second row, KE = 220500 J

Third row, KE = 183750 J

Fourth row, KE = 205800 J

That's also not paying any attention to significant digits because your velocity only had 1 and that's not enough to do the problem justice. I left all the digits in the answer. Round how your teacher tells you to.

3 0
2 years ago
When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s
d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(2)^2 = 0.5 J

2) when speed is 3 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(3)^2 = 1.125 J

3) when speed is 4 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(4)^2 = 2 J

4) when speed is 5 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(5)^2 = 3.125 J

5) when speed is 6 m/s

kinetic energy is given as

K = \frac{1}{2}mv^2

K = \frac{1}{2}(0.250)(6)^2 = 4.5 J

3 0
3 years ago
Read 2 more answers
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