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saw5 [17]
3 years ago
8

Which statement is most likely correct?

Physics
2 answers:
adoni [48]3 years ago
7 0

The 3rd statement is most probable.

Len [333]3 years ago
6 0

it's either the first or second choice

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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
A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along
Pachacha [2.7K]

Answer:

The work done by a particle from x = 0 to x = 2 m is 20 J.

Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

F(x)=(3\ N/m^2)x^2+(6\ N/m)x

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

W=\int\limits {F{\cdot} dx}

W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J

So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
Is the magnetic field inside a toroid uniform?
victus00 [196]
If I am not wrong i thinks it is in the toroid uniforms
4 0
3 years ago
An electric field of intensity 3.80 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.
lakkis [162]

Answer: 1.55 x 10⁴ Nm²c^-1

Explanation: The electric flux, electric field intensity and area are related by the formulae below.

Φ= EAcosθ,

Where Φ= electric flux (Nm²c^-1)

E =electric field intensity (N/m²)

A = Area (m²)

θ= this is angle between the planar area and the magnetic flux

For our question E=3.80KN/c= 3800 N/c

A= 0.700 x 0.350= 0.245m²

θ= 0° ( this is because the electric field was applied along the x axis, thus the electric flux will be parallel to the area).

Hence Φ= 3800 x 0.245 x cos(0)

= 3800 x 0.245 x 1 (value of cos 0° =1)

= 1.55 x 10⁴ Nm²c^-1

Thus the electric field is 1.55 x 10⁴ Nm²c^-1

4 0
3 years ago
You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.5 m. a) What fraction of its initial energy is lost
tangare [24]
The fraction of energy that is lost is 25%, it depends how fast the ball was going until it lost 25% of its energy, the gravitational energy was transferred into the kinetic energy that helped the ball bounce back
4 0
3 years ago
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