6- A metal block measures 10 cm x 2 cm x 2 cm. what is its volume? how many blocks each

A cyclist overcomes a resistive force of 30 N in order to cycle 30 m. It takes her 6 seconds to cycle this distance. Calculate t

zmey [24]

Answer:

Power = 15[W]

Explanation:

This problem can be solved using the work definition.

Work is equal to the product of the force by the distance, for this problem we have:

F = force = 30 [N]

d = distance = 30 [m]

w = work = F * d = 30*30 = 900 [J], "units in joules"

The power is defined as the work done in an interval of time.

P = w / t

where:

t = time [s]

therefore

P = w / t

P = 900 / 6

P = 150 [W] "units in watts"

6 0

3 years ago

The primary of a step-up transformer is connected across the terminals of a standard wall socket, and resistor 1 with a resistan

alexira [117]

The turns ratio is the factor that determines voltage andcurrent. In order to have the same current across the resistorin the primary as the resistor in the secondary, then:--N(p) = Primary turnsN(s) = Secondary turnsR(2) = Primary resistorR(1) = Secondary resistor--R(2)/R(1) = N(p)/N(s)R(2) = R(1)*(N(p)/N(s))--If arbitrary values are plugged in, you will see that this step up transformer will require 2x the resistance required in the secondary, R(1), to obtain the same current. Thus R(2) will be 1/2 the value of R(1). This is due to the stepped up voltage in the secondary.

3 0

3 years ago

a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100

Anastaziya [24]

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

let's reduce the magnitudes to the SI system

r = 10 cm = 0.10 m

m = 10 g = 0.010 kg

A = 100 cm² = 0.01 m²

the definition of density is

ρ = m / V

the volume of a sphere

V = $\frac{4}{3} \ \pi r^{3}$

V = $\frac{4}{3}$ π 0.1³

V = 4.189 10⁻³ m³

let's calculate the density of the ball

ρ = $\frac{0.010}{4.189 \ 10^{-3} }$

ρ = 2.387 kg / m³

the tabulated density of water is

ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

B - W = 0

B = W

where B is the thrust that is given by Archimedes' principle

ρ_liquid g V_submerged = m g

V_submerged = m / ρ_liquid

we calculate

V _submerged = 0.10 9.8 / 997

V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

V = A h

h = V / A

let's calculate

h = 9.83 10⁻⁴ / 0.01

h = 0.0983 m

this is equal to h = 9.83 cm

8 0

2 years ago

The temperature of air changes from 0 to 10°C while its velocity changes from zero to a final velocity, and its elevation change

aliya0001 [1]

Answer:

Final velocity = 119.83 m/s

Final elevation = 731.9 m

Explanation:

We are told that temperature of air changes from 0 to 10°C

Thus;

Change in temperature; ΔT = 10 - 0 = 10°C

Also, its velocity changes from zero to a final velocity. Thus;

v1 = 0 m/s

v2 is unknown

Also, its elevation changes from zero to a final elevation.

So, z1 = 0 and z2 is unknown

Now, we want to find v2 and z2 when the internal, kinetic and potential energy are equal.

Thus Equating the formula for both kinetic and internal energy gives;

½m(v2² - v1²) = mc_v•ΔT

m will cancel out and v1 is zero to give;

v2² = 2c_v•ΔT

v2 = √(2c_v•ΔT)

Where c_v is specific heat of constant volume of air with a constant value of 718 J/Kg.K

Thus;

v2 = √(2 × 718 × 10)

v2 = √14360

v2 = 119.83 m/s

To find z2, we will equate potential energy formula to that of the internal energy.

Thus;

mg(z2 - z1) = mc_v•ΔT

m will cancel out and since z1 is zero, then we have;

z2 = (c_v•ΔT)/g

z2 = 718 × 10/9.81

z2 = 731.9 m

4 0

3 years ago

The figure shows two springs (k1 = 10 N/m and k2 = 20 N/m ) attached to a block that can slide on a frictionless surface. In the

Rainbow [258]

Answer:

Explanation:

a )

Energy stored by left spring when compressed = 1/2 k x²

= .5 x 10 x .02² = .002 J .

Let compression in right spring = y

energy stored to right spring = 1/2 k y²

1/2 k y² = 0.002

.5 x 20 x y² = 0.002

y = .01414 m

= 1.4 cm

7 0

2 years ago