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3 years ago

A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the

Physics

1 answer:

Tems11 [23]3 years ago

6 0

Mass=m=20kg
Radius=r=4m
Velocity=10m/s=v

We know

$\boxed{\sf F_c=\dfrac{mv^2}{r}}$

$\\ \sf\longmapsto F_c=\dfrac{20(10)^2}{4}$

$\\ \sf\longmapsto F_c=\dfrac{20(100)}{4}$

$\\ \sf\longmapsto F_c=\dfrac{2000}{4}$

$\\ \sf\longmapsto F_c=500N$

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Can someone please help me?? i’ll give brainilist

Ksivusya [100]

Probably 90 j but im not sure I haven’t done any work like this in a while

7 0

3 years ago

A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 14.413

natima [27]

Answer:

The value of charge q₃ is 40.46 μC.

Explanation:

Given that.

Magnitude of net force $F=14.413\ N$

Suppose a point charge q₁ = -3 μC is located at the origin of a co-ordinate system. Another point charge q₂ = 7.7 μC is located along the x-axis at a distance x₂ = 8.2 cm from q₁. Charge q₂ is displaced a distance y₂ = 3.1 cm in the positive y-direction.

We need to calculate the distance

Using Pythagorean theorem

$r=\sqrt{x_{2}^2+y_{2}^2}$

Put the value into the formula

$r=\sqrt{(8.2\times10^{-2})^2+(3.1\times10^{-2})^2}$

$r=0.0876\ m$

We need to calculate the magnitude of the charge q₃

Using formula of net force

$F_{12}=kq_{2}(\dfrac{q_{3}}{r_{3}^2}+\dfrac{q_{1}}{r_{1}^2})$

Put the value into the formula

$14.413=9\times10^{9}\times7.7\times10^{-6}(\dfrac{q_{3}}{(0.0438)^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})$

$(\dfrac{q_{3}}{(4.38\times10^{-2})^2}+\dfrac{-3\times10^{-6}}{(0.0876)^2})=\dfrac{14.413}{9\times10^{9}\times7.7\times10^{-6}}$

$\dfrac{q_{3}}{(0.0438)^2}=207\times10^{-4}+3.909\times10^{-4}$

$q_{3}=0.0210909\times(0.0438)^2$

$q_{3}=40.46\times10^{-6}\ C$

$q_{3}=40.46\ \mu C$

Hence, The value of charge q₃ is 40.46 μC.

5 0

3 years ago

A hollow sphere of inner radius 8.82 cm and outer radius 9.91 cm floats half-submerged in a liquid of density 948.00 kg/m^3. (a)

kari74 [83]

Answer:

a) 0.568 kg

b) 474 kg/m³

Explanation:

Given:

Inner radius = 8.82 cm = 0.0882 m

Outer radius = 9.91 cm = 0.0991 m

Density of the liquid = 948.00 Kg/m³

a) The volume of the sphere = $\frac{4\pi}{3}\times(0.0991^2-0.0882^2)$

volume of sphere = 0.0012 m³

also, volume of half sphere = $\frac{\textup{Total volume}}{\textup{2}}$

volume of half sphere = $\frac{\textup{0.0012}}{\textup{2}}$

Volume of half sphere =0.0006 m³

Now, from the Archimedes principle

Mass of the sphere = Weight of the volume of object submerged

Mass of the sphere = 0.0006× 948.00 = 0.568 kg

b) Now, density = $\frac{\textup{Mass}}{\textup{Volume}}$

Density = $\frac{\textup{0.568}}{\textup{0.0012}}$

Density = 474 kg/m³

8 0

3 years ago

Which substance is a greenhouse gas that is emitted by human activities, is the most widespread, and remains in the air the long

e-lub [12.9K]

Answer:

I am not really sure, but it is probably Carbon Dioxide

Explanation:

3 0

4 years ago

Read 2 more answers

A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine

Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = $\frac{4}{3}\pi(\frac{d}{2})^3$

Volume of the balloon = $\frac{4}{3}\pi(\frac{0.296}{2})^3$

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = $\frac{3}{2}\times K_BT$

where,

Boltzmann constant, $K_B=1.3807\times10^{-23}J/K$

Average kinetic energy = $\frac{3}{2}\times1.3807\times10^{-23}\times292$

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = $\frac{3RT}{m}$

where, m is the molar mass of the Helium = 0.004 Kg

rms speed = $\frac{3\times8.314\times292}{0.004}$

rms speed = 1349.35 m/s

5 0

3 years ago