Calculate the power of a car that does 50,000 J of work in 7s during a race

Calculate the weight of a 1 kg mass at earth's surface. The mass of the of the Earth's surface if the mass of the earth is 6 x 1

yuradex [85]

Answer:

9.8N

Explanation:

Here we can get gravitational acceleration according to the place where object is placed by bellow equation

g = GM/R²

g - Gravitational Acceleration

G - Gravitational constant (6.67×10-11)

R - Distance ( Radius )

g = 6.67 × 10-11 × 1024 /(6.37×106)²

g = 9.8 m/s²

There for

Weight = Mass × Gravitational acceleration

= 1×9.8

= 9.8 N

4 0

3 years ago

Jane starts from her house to take a stroll in her neighborhood. After walking for 2 hours at a steady pace, she has walked 4 mi

9966 [12]

Answer:

C. 2 mph

Explanation:

speed equals total distance covered in total time of of travel

since,

total distance = 4 miles

time taken = 2 hrs

speed = 4m / 2hrs =2mph

3 0

4 years ago

Read 2 more answers

You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle

nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for 90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A