All categories

2 years ago

If an object has a mass of 20 kg, what is the force of gravity acting on it on earth? A. 32.67 N B. 2.04 kg C. 1.96 kg D. 196 N

Physics

2 answers:

Brilliant_brown [7]2 years ago

7 0

M=20 kg
g on earth=9.8m sec^-1
F=m*a
F=20*9.8
F=196N

Nimfa-mama [501]2 years ago

4 0

Answer:

D. 196 N

Explanation:

The force of gravity acting on an object located at the Earth's surface is given by:

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem:

- m=20 kg is the mass of the object

- g = 9.81 m/s^2 is the acceleration due to gravity at Earth's surface

Substituting these data into the formula, we get

$F=(20 kg)(9.81 m/s^2)=196.2 N \sim 196 N$

You might be interested in

A flat square plate of side 20cm moves over other similar plate with a thin layer of 0.4cm of a liquid between them with force 1

antiseptic1488 [7]

Answer:

$\mu=0.98\ Pa.s$

Explanation:

Given:

dimension of square plate, $l=0.2\ m$

thickness of fluid layer, $dy=0.004\ m$

force on the fluid due to plate, $F=1\ kgw=1\times 9.8=9.8\ N$

velocity of plate, $du=1\ m.s^{-1}$

<u>Using Newton's law of viscosity:</u>

$\tau=\mu.\frac{du}{dy}$ ..........................................(1)

where:

$\tau=$ shear force on the surface on the fluid

$\mu=$ coefficient of (dynamic) viscosity

Now, shear force:

$\tau=\frac{shear\ force}{area}$

$\tau=\frac{9.8}{0.2\times0.2} \ Pa$

Putting respective values in eq.(1)

$\frac{9.8}{0.2\times0.2}=\mu\times\frac{1}{0.004}$

$\mu=0.98\ Pa.s$

7 0

3 years ago

The 64.5-kg climber in is supported in the “chimney” by the friction forces exerted on his shoes and back. The static coefficien

kobusy [5.1K]

Answer:

The minimum force the climber must exert is about 439N.

Explanation:

We use the relationship between friction and normal force to answer this question:

$F_{friction} = \mu_{static} \cdot F_{normal}\implies F_{normal}=\frac{F_{friction}}{\mu_{static}}$

We are given the static coefficients of friction but need to determine the friction force. To do that we consider the totality of forces acting on this hapless gentleman stuck in a chimney. There is the gravity acting downward (+), then there are two friction forces acting upward (-), namely through his shoes and his back. The horizontal force exerted by the climber on both walls of the chimney is the same and is met with equally opposing normal force. Since the climber is not falling the net force in the vertical direction is zero:

$F_{net} = 0 = F_g - F_{shoes}-F_{back}= mg - \mu_{shoes}F_{norm}-\mu_{back}F_{norm}\\F_{norm}=\frac{mg}{\mu_{shoes}+\mu_{back}}=\frac{64.5kg\cdot 9.8\frac{m}{s^2}}{0.8+0.64}\approx 438.96N\\$

The normal force in this equilibrium is about 439N and because we are told that the static friction forces are both at their maximum, this value is at the same time the <em>minimum</em> force needed for the climber to avoid starting slipping down the chimney.

7 0

2 years ago

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

$emf=5.4090\times 10^{-4}V$

7 0

3 years ago

In economics what goods and services will be produced?

poizon [28]

Answer:A market economy is a system where the laws of supply and those of demand direct the production of goods and services. 1 Supply includes natural resources, capital, and labor. Demand includes purchases by consumers, businesses, and the government. Businesses sell their wares at the highest price consumers will pay.

Explanation:

5 0

2 years ago

A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnet

Fiesta28 [93]

Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

Explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

$a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C$

Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C

6 0

2 years ago