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2 years ago

15

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.6 m and the horizontal

range of the ball from the base of the platform is 20 m. What is the horizontal velocity of the ball just before it touches the ground?

1 answer:

Sever21 [200]2 years ago

3 0

Explanation:

the answer is in the above image

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How do I solve such problem???

labwork [276]

1). Calculate how long it takes an object to fall 4,000 m after it's dropped. (Use D = (1/2) (g) (T²) . D is 4,000 m. g = 9.8 m/s². Find T .)

2). Calculate how far the object will move HORIZONTALLY in that length of time, if it's moving at 75 m/s. (Distance = (75 m/s) x (time) . )

6 0

3 years ago

24 A uniform electric field of magnitude 1.1×104 N/C is perpendicular to a square sheet with sides 2.0 m long. What is the elect

Tatiana [17]

Answer:

$44,000 Nm^2/C$

Explanation:

The electric flux through a certain surface is given by (for a uniform field):

$\Phi = EA cos \theta$

where:

E is the magnitude of the electric field

A is the area of the surface

$\theta$ is the angle between the direction of the field and of the normal to the surface

In this problem, we have:

$E=1.1\cdot 10^4 N/C$ is the electric field

L = 2.0 m is the side of the sheet, so the area is

$A=L^2=(2.0)^2=4.0 m^2$

$\theta=0^{\circ}$ , since the electric field is perpendicular to the surface

Therefore, the electric flux is

$\Phi =(1.1\cdot 10^4)(4.0)(cos 0^{\circ})=44,000 Nm^2/C$

4 0

2 years ago

In which condition the acceleration of a moving vehicle become zero

Alekssandra [29.7K]

Explanation:

When,the vehicle has uniform velocity, it's acceleration becomes zero

4 0

2 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for <em>M</em>, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

5 0

2 years ago

A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.

denpristay [2]

Answer:

The mass of the rule is 56.41 g

Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0 22cm 100cm

-------------------------Δ------------------------------------

↓ ↓

200g m₂

Apply principle of moment

(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ = (200 g)(22 cm) / (78 cm)

m₂ = 56.41 g

Therefore, the mass of the rule is 56.41 g

3 0

3 years ago