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3 years ago

an object weighs 98 n on earth. How much does it weigh on planet x where the acceleration due to gravity in 6 m/s^2

2 answers:

7 0

60 N because 98N=mg (here g= 9.8 on earth) thus mass can be calculated which is 98/9.8 = 10kg Now,new weight with g = 6m/s^2 =m×g' (here g' is new acceleration of the new planet) = 10×6=60N

Ainat [17]3 years ago

6 0

Answer:

The weight on planet x is 60 N

Explanation:

It is given that,

Weight of an object on earth, W = 98 N

Value of acceleration due to gravity, $g=9.8\ m/s^2$

Weight, $W=mg$

$m=\dfrac{W}{g}=\dfrac{98\ N}{9.8\ m/s^2}=10\ kg$

Since, mass of an object remains same everywhere.

If the value of acceleration due to gravity on planet x is, $a=6\ m/s^2$

So, weight on the planet x is,

$W=m\times a$

$W=10\ kg\times 6\ m/s^2$

W = 60 N

So, the weight on planet x is 60 N. Hence, this is the required solution.

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Answer:

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3 years ago

The distance between 4 nodes (3 sections, 2 sections= wavelength) is 15.0cm. The frequency of the source is 10Hz. What's the spe

Natalija [7]

Answer:

the speed of the waves is 150 cm/s

Explanation:

Given;

frequency of the wave, f = 10 Hz = 10

distance between 4 nodes, L = 15.0 cm

The wavelength (λ) of the wave is calculated as follows;

Node to Node = λ/2

L = 2(Node to Node) = (4 Nodes) = 2 (λ/2) = λ

Thus, λ = L = 15.0 cm

The speed (v) of the wave is calculated as follows;

v = fλ

v = 10 Hz x 15.0 cm

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Therefore, the speed of the waves is 150 cm/s

7 0

2 years ago

Nresistors, each having resistance equal to 1 2, are arranged in a circuit first in series and then in parallel. What is the rat

aleksandr82 [10.1K]

Answer:

option (b)

Explanation:

Let the resistance of each resistor is R.

In series combination,

The effective resistance is Rs.

rs = r + R + R + .... + n times = NR

Let V be the source of potential difference.

Power in series

Ps = v^2 / Rs = V^2 / NR ..... (1)

In parallel combination

the effective resistance is Rp

1 / Rp = 1 / R + 1 / R + .... + N times

1 / Rp = N / R

Rp = R / N

Power is parallel

Rp = v^2 / Rp = N V^2 / R ..... (2)

Divide equation (1) by equation (2) we get

Ps / Pp = 1 / N^2

5 0

3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

Earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

kobusy [5.1K]

The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x $10^{24}$ Kg

<h3>Relationship between Linear and angular speed</h3>

Linear speed is the product of angular speed and the maximum displacement of the particle. That is,

V = Wr

Where

V = Linear speed

W = Angular speed

r = Radius

Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed

W = 2 $\pi$ /T

W = (2 x 3.143) / (365.26 x 24)

W = 6.283 / 876624

W = 7.2 x $10^{-4}$ Rad/hr

The Earth’s average orbital speed V = Wr

V = 7.2 x $10^{-4}$ x 149.6 x $10^{6}$

V = 107225.5 kilometers per hours.

b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law

M = (4 $\pi ^{2}$ $r^{3}$ ) / G $T^{2}$

M = (4 x 9.8696 x 3.35 x $10^{24}$ ) / (6.67 x $10^{-11}$ x 7.68 x $10^{11}$ <em>)</em>

<em>M = 1.32 x </em> $10^{26}$ / 51.226

M = 2.58 x $10^{24}$ Kg

Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x $10^{24}$ Kg

Learn more about Orbital Speed here: brainly.com/question/22247460

#SPJ1

3 0

1 year ago