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A block with a mass of 9.00 kg is pulled at a constant speed across a horizontal tabletop with a spring scale. The scale reads 6

snow_tiger [21]

Answer:0.69

Explanation:

Coefficient of kinetic friction=f/R=61.8/90=0.69

7 0

3 years ago

The bae chart shows the percentage of adults in the U.S. living with certain heart disease risk factors. Which conclusion can be

d1i1m1o1n [39]

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6 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

A calorimeter is a container that is insulated from the outside, so a negligible amount of energy enters or leaves the container

Len [333]

Answer:

This is a heat balance question with negligible heat losses to the surroundings.

Sum of heat losses = Sum of the heat gains

We are required to find the final temperature of the calorimeter system T1?

we were given

C = specific heat capacities of Copper c, Water w, Ice i and Lead l

m = masses as mc = 0.1kg, mw = 0.16kg, mi=0.018kg, ml=0.75kg,

T = Temperature of the components, Lead Tl= 225C and at ambient pressure, Temp. To of the water, ice and copper = 25C

Since, at thermal equilibrium after adding the Lead, the lead temperature Tl decreases while copper, water and ice temp decrease, so we have respective heat losses 'q' as,

qc + qw + qi + ql =0

which means that

(mc x Cc x ΔTC) + (mw x Cw x ΔTC) + (mi x Ci x ΔTC) + (ml x Cl x ΔTC) = 0

(0.1x390x (T1 - 25)) + (0.16x4190x (T1 - 25)) + (0.018x4190x (T1 - 0)) + (0.750x130x (T1 - 225)) = 0

882.32T1 - 41558 = 0

T1 = 47.1K

Hence, the final temperature of the calorimeter system is 47.1K

3 0

3 years ago

What is the vector product of two vectors????

Nadya [2.5K]

Answer:

The magnitude of the vector product of two vectors can be constructed by taking the product of the magnitudes of the vectors times the sine of the angle (<180 degrees) between them. The magnitude of the vector product can be expressed in the form: and the direction is given by the right-hand rule.

7 0

3 years ago