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valentinak56 [21]
2 years ago
8

If the force of gravity between a book of mass 0.50 kg and a calculator of 0.100 kg is 1.5 × 10-10 N, how far apart are they?  (

___/ 2)
Physics
1 answer:
valkas [14]2 years ago
8 0
The gravitational force between two masses m₁ and m₂ is
F=G \frac{m_{1} m_{2}}{d^{2}}
where
G = 6.67408 x 10⁻¹¹ m³/(kg-s²), the gravitational constant
d =  distance between the masses.

Given:
F = 1.5 x 10⁻¹⁰ N
m₁ = 0.50 kg
m₂ = 0.1 kg

Therefore
1.5 x 10⁻¹⁰ N = (6.67408 x 10⁻¹¹ m³/(kg-s²))*[(0.5*0.1)/(d m)²]
d² = [(6.67408x10⁻¹¹)*(0.5*0.1)]/1.5x10⁻¹⁰
     = 0.0222
d = 0.1492 m = 149.2 mm

Answer: 149.2 mm
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A celebrating student throws a water balloon horizontally from a dormitory window that is 50 m above the ground. It hits the gro
Contact [7]

Answer:

a) The horizontal velocity of the balloon just before it hits the ground is 6 m/s

b) The magnitude of the vertical velocity of the balloon just before it hits the ground is 98 m/s.

Explanation:

Hi there!

The velocity and position vectors of the water balloon are given by the following equations:

r =(x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v =(v0x, v0y + g · t)

where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward position as positive) .

v = velocity vector at time t.

a) Please, see the attached figure for a graphic description of the problem.

Considering the origin of the frame of reference as the point of launch, notice that the position vector when the balloon hits the ground is

r1 = (60, -50) m

Then:

r1x = 60 m = v0x · t

r1y = -50 m = 1/2 · (-9.8 m/s²) · t²

(notice that the initial vertical velocity is zero, see figure).

Solving r1y for t:

(-50 m · 2) / -9.8 m/s² = t²

t = 10 s

Now, let´s replace t in the r1x equation and solve it for the horizontal component of the velocity:

60 m = v0x · 10 s

v0x = 60 m / 10 s

v0x = 6 m/s

The initial horizontal component of the velocity is 6 m/s. This velocity is constant because there is no air resistance. Then, just before the balloon hits the ground, it will have a horizontal velocity of 6 m/s.

b) To calculate the vertical component of the velocity when the balloon hits the ground, let´s use the equation of the vertical component of the velocity:

v1y = v0y + g · t

Since v0y = 0

v1y = -9.8 m/s² · (10 s) = -98 m/s

The magnitude of the vertical velocity of the balloon when it hits the ground is 98 m/s.

4 0
2 years ago
Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0
2 years ago
A solid block of mass m is suspended in a liquid by a thread. The density of the block is greater than that of the liquid. Initi
Tanzania [10]

Answer:

(C) T

The tension T at equilibrium will be equal to the Buoyant force.

The Buoyant force is given by:

Fb = density x acceleration due to gravity x volume displaced

The change in height doesn't affect the Buoyant force and hence the tension.

Note: The figure of question is added in the attachment

3 0
2 years ago
A skydiver is using wind to land on a target that is 50 m away horizontally. The skydiver starts from a height of 70 m and is fa
elena55 [62]

Answer:

Answer:

15.67 seconds

Explanation:

Using first equation of Motion

Final Velocity= Initial Velocity + (Acceleration * Time)  

v= u + at

v=3

u=50

a= - 4 (negative acceleration or deceleration)  

3= 50 +( -4 * t)

-47/-4 = t

Time = 15.67 seconds

6 0
2 years ago
A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle wi
Kamila [148]

Answer:

A.)1.52cm

B.)1.18cm

Explanation:

angular speed of 120 rev/min.

cross sectional area=0.14cm²

mass=12kg

F=120±12ω²r

=120±12(120×2π/60)^2 ×0.50

=828N or 1068N

To calculate the elongation of the wire for lowest and highest point

δ=F/A

= 1068/0.5

δ=2136MPa

'E' which is the modulus of elasticity for alluminium is 70000MPa

δ=ξl=φl/E =2136×50/70000=1.52cm

δ=F/A=828/0.5

=1656MPa

δ=ξl=φl/E

=1656×50/70000=1.18cm

δ=1.18cm

6 0
2 years ago
Read 2 more answers
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