A runner makes one lap around a 200m track of a time of 25s . What were the runner's (a) average speed

Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,

Nikolay [14]

Explanation:

It is given that,

Power consumed by toaster oven, $P=1.4\ kW$

Time taken, t = 6 minutes = 0.1 hour

Energy used by toaster, $W_{toaster}=P\times t$

$W_{toaster}=1.4\ kW\times 0.1\ h=0.14\ kWh$

Similarly,

Power consumed by toaster oven, $P=11\ W$

Time taken, t = 9 minutes = 0.15 hour

Energy used by toaster, $W_{toaster}=P\times t$

$W_{toaster}=11\ W\times 0.15\ h=1.65\ Wh=0.0016\ kWh$

Hence, this is the required solution.

3 0

3 years ago

Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?

Talja [164]

Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

$\frac{10N}{50kg}=a$

A newton is just $\frac{kg·m}{s^{2}}$

$a=\frac{\frac{10kg·m}{s^{2}}}{50kg}$

The s^2 and 50 kg you multiply

$a=\frac{10kg·m}{50kg·s^{2}}$

The kg's cancel and 10/50 is 1/5

$\frac{1}{5}·\frac{m}{s^{2}}$

So the acceleration is 1/5 m/s^2

3 0

3 years ago

Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t

ruslelena [56]

Answer:

$x_1= 0.0425m$

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

$m*g = k*x$

With two springs, let, T1 be the tension in each spring, x1 be the extension of each spring. The spring constant of each spring is 2k so:

$T_1 = 2k*x_1$

$2T_1 = m*g=4k x_1$

Solve to x1

$x_1=\frac{m*g}{4k}$

$x_1=\frac{k*x}{4*k}$

$x_1=\frac{x}{4}$

$x_1 = 0.170 / 4$

$x_1= 0.0425m$

7 0

3 years ago

The area of a rectangular park is 4 mi^2. The park has a width that is equal to "w", and a length that is 3 mi longer than the w

goldfiish [28.3K]

Answer:

l= 4 mi : width of the park

w= 1 mi : length of the park

Explanation:

Formula to find the area of the rectangle:

A= w*l Formula(1)

Where,

A is the area of the rectangle in mi²

w is the width of the rectangle in mi

l is the width of the rectangle in mi

Known data

A = 4 mi²

l = (w+3)mi Equation (1)

Problem development

We replace the data in the formula (1)

A= w*l

4 = w* (w+3)

4= w²+3w

w²+3w-4= 0

We factor the equation:

We look for two numbers whose sum is 3 and whose multiplication is -4

(w-1)(w+4) = 0 Equation (2)

The values of w for which the equation (2) is zero are:

w = 1 and w = -4

We take the positive value w = 1 because w is a dimension and cannot be negative.

w = 1 mi :width of the park

We replace w = 1 mi in the equation (1) to calculate the length of the park:

l= (w+3) mi

l= ( 1+3) mi

l= 4 mi

8 0

4 years ago

An Atwood's machine consists of two different masses, both hanging vertically and connected by an ideal string which passes over

Tasya [4]

Answer:

V₁ = √ (gy / 3)

Explanation:

For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2

Starting point

Em₀ = U₁ + U₂

Em₀ = m₁ g y₁ + m₂ g y₂

Let's place the reference system at the point where the mass m1 is

y₁ = 0

y₂ = y

Em₀ = m₂ g y = 2 m₁ g y

End point, at height yf = y / 2

$E_{mf}$ = K₁ + U₁ + K₂ + U₂

$E_{mf}$ = ½ m₁ v₁² + ½ m₂ v₂² + m₁ g $y_{f}$ + m₂ g $y_{f}$

Since the masses are joined by a rope, they must have the same speed

$E_{mf}$ = ½ (m₁ + m₂) v₁² + (m₁ + m₂) g $y_{f}$

$E_{mf}$ = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

How energy is conserved

Em₀ = $E_{mf}$

2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g $y_{f}$

2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2

3/2 v₁² = 2 g y -3/2 g y

3/2 v₁² = ½ g y

V₁ = √ (gy / 3)

5 0

3 years ago