What unit of measurement would be used to measure the weight of a book

A mechanic test driving a car that she has just given a tune-up accelerates from rest to 50.0 m/s in 9.8 s. How far (in meters)

Stels [109]

Answer:

245 m

Explanation:

v = at + v₀

50.0 m/s = a (9.8 s) + 0 m/s

a = 5.10 m/s²

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (9.8 s) + ½ (5.10 m/s²) (9.8 s)²

x = 245 m

6 0

3 years ago

An object with a mass of 70 kilograms is supported at a height 8 meters above the ground. What's the potential energy of the obj

alexira [117]

Ep= mgh

70 x 9.8 x 8

Ep= 5,488J

5 0

2 years ago

Right or left? Summer and Winter or day and night?

evablogger [386]

Answer:

The answer is left

Explanation:

Sorry if im worng

6 0

3 years ago

Read 2 more answers

If the depth of water in a well is 10m, what is the pressure exerted by it the bottom of the well ? ( Use g = 10 m/s2)

Tasya [4]

Answer:

The precise answer depends on the density and therefore the temperature of the water, but we can obtain a reasonable approximation by assuming that the density of the water is 1000 kilograms per cubic meter (kg/m³).

Since the depth of the water in the well is 10 m, the volume of water directly above an area A of a square meters (m²) at the bottom of the well is 10×a m³.

Since the density of the water is 1,000 kg/m³, the mass of water directly above area A is (1,000 kg/m³) × (10×a m³) = (1000×10×a kg) = 10,000×a kg.

Since g = 9.8 m/s², the force of gravity acting on the water directly above area A is (9.8 m/s²) × (10,000×a kg) = 9.8×10,000×a N (newtons) = 98,000×a N.

So the pressure of water acting on area A is (98,000×a N)/(a m²) = (98,000×a)/a N/m² = 98,000 pascals (pa). And since A could be any given area at the bottom of the well, this is the pressure at any point at the bottom of the well.

So the pressure at the bottom of the well is 98,000 pascals (or 98,000/101,325 standard atmospheres = 560/579 atmospheres ~ 0.967 standard atmospheres).

Please comment below if you have any questions.

7 0

3 years ago

f a single circular loop of wire carries a current of 45 A and produces a magnetic field at its center with a magnitude of 1.50

Lelu [443]

Answer:

Radius of the loop is 0.18 m or 18 cm

Explanation:

Given :

Current flowing through the wire, I = 45 A

Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T

Number of turns in circular wire, N = 1

Consider R be the radius of the circular wire.

The magnetic field at the center of the current carrying circular wire is determine by the relation:

$B=\frac{N\mu_{0} I}{2R}$

Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.

Substitute the suitable values in the above equation.

$1.50\times10^{-4} =\frac{4\pi \times10^{-7}\times45 }{2R}$

R = 0.18 m

4 0

2 years ago