Ask question

Ask question

All categories

2 years ago

4

A 10kg object is near a planet’s surface such that the gravitational field strength is 4Nkg. With what force is the planet attra

cted to the 10kg object?
A-4 Nkg

b-10N

c- 40N

d-100N

1 answer:

dsp732 years ago

7 0

Answer:

C. 40 N

Explanation:

Gravitational field strength = force/mass.

Force = mass* gravitational field strength.

F = 10*4 = 40 N.

You might be interested in

A sailboat starts from rest and accelerates at a rate of 0.21 m/s^2 over a distance of 280 m. find the magnitude of the boat's f

sasho [114]

We use the kinematic equations,

$v=u+at$ (A)

$S= ut + \frac{1}{2} at^2$ (B)

Here, u is initial velocity, v is final velocity, a is acceleration and t is time.

Given, $u=0$ , $a=0.21 \ m/s^2$ and $s= 280 m$ .

Substituting these values in equation (B), we get

$280 \ m = 0 +\frac{1}{2} (0.21 m/s^2) t^2 \\\\ t^2 = \frac{280 \times 2}{0.21 } \\\\ t= 51.63 \ s$ .

Therefore from equation (A),

$v = 0 + (0.21) \times (51.63 s)= 10.84 \ m/s$

Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s

8 0

2 years ago

Ricardo and Jane are standing under a tree in the middle of a pasture. An argument ensues, and they walk away in different direc

Advocard [28]

Answer:

the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south

Explanation:

given information:

Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus

A= 27

Ax = 27 sin 60 = - 23.4

Ay = 27 cos 60 = 13.5

Jane walks 16.0 m in a direction 30.0 ∘ south of west, so

B = 16

Bx = 16 cos 30 = -13.9

By = 16 sin 30 = -8

the direction that should be walked by Ricardo to go directly to Jane

R = √A²+B² - (2ABcos60)

= √27²+16² - (2(27)(16)(cos 60))

= 23.52 m

now we can use the sines law to find the angle

tan θ = $\frac{R_{y} }{R_{x} }$

= By - Ay/Bx -Ax

= (-8 - 13.5)/(-13.9 - (-23.4))

θ = 90 - (-8 - 13.5)/(-13.9 - (-23.4))

= 24° east of south

4 0

2 years ago

A thin film of polystyrene is used as an antireflective coating for fabulite (known as the substrate). the index of refraction o

kvasek [131]

To solve this problem, we assume that the wavelength of the light in air is 500 nanometers.

For this case we only need the refractive index of the polystyrene. For an antireflective coating, we need a quarter of wave thickness at the wavelength in the air. Which means that the antireflective coating needs to be as thick as 1/4 of the wavelength, divided by the coating’s refractive index. This is expressed mathematically in the form:

x = λ / (4 * n)

where,

x = thickness

λ = wavelength of light

n = index of refraction of polystyrene

Substituting:

x = 500 nm / (4 * 1.49)
x = 500 nm / 5.96
x = 83.90 nm

6 0

2 years ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

2 years ago

(8%) Problem 9: Helium is a very important element for both industrial and research applications. In its gas form it can be used

exis [7]

Answer:

2046.37 kPa

Explanation:

Given:

Number of moles, n = 125

Temperature, T = 20° C = 20 + 273 = 293 K

Radius of the cylinder, r = 17 cm = 0.17 m

Height of the cylinder, h = 1.64 m

thus,

volume of the cylinder, V = πr²h

= π × 0.17² × 1.64

= 0.148 m³

Now,

From the ideal gas law

we have

PV = nRT

here,

P is the pressure

R is the ideal gas constant = 8.314 J / mol. K

thus,

P × 0.148 = 125 × 8.314 × 293

or

P × 0.148 = 304500.25

or

P = 2046372.64 Pa = 2046.37 kPa

6 0

2 years ago