What makes the world go round?

How much force is needed to accelerate a 66kg skier at 2m/s

nika2105 [10]

264..forces is required

5 0

3 years ago

A 32.5 g cube of aluminum initially at 45.8 °C is submerged into 105.3 g of water at 15.4 °C. What is the final temperature of b

lbvjy [14]

Answer:

$T = 17.26 ^oC$

Explanation:

At thermal equilibrium we have heat given by aluminium must be equal to the heat absorbed by the water

so we will have

$Q_1 = Q_2$

$m_1s_1\Delta T_1 = m_2s_2\Delta T_2$

so we will have

$32.5(900)(45.8 - T) = 105.3(4186)(T - 15.4)$

so we have

$(45.8 - T) = 15.1(T - 15.4)$

so we have

$16.1 T = 277.87$

$T = 17.26 ^oC$

3 0

3 years ago

What is the period of oscillation of a mass 40kg on a spring with constant k=10N/m?

iVinArrow [24]

The period of oscillation of the system is 12.56 s

Explanation:

The period of oscillation of a spring-mass system is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where

m is the mass attached to the spring

k is the spring constant

For the system in this problem, we have

m = 40 kg

k = 10 N/m

Substituting into the equation, we find

$T=2\pi \sqrt{\frac{40}{10}}=12.56 s$

#LearnwithBrainly

7 0

3 years ago

Two microwave signals of nearly equal wavelengths can generate a beat frequency if both are directed onto the same microwave det

alekssr [168]

Answer:

1.5106 cm

Explanation:

The beat frequency is equal to the absolute value of the difference between the frequencies of the two signals:

$f_B = |f_1 - f_2|$

using the wave equation, we can re-write each frequency as

$f=\frac{c}{\lambda}$

where c is the speed of light and $\lambda$ is the wavelength. Therefore,

$f_B = |\frac{c}{\lambda_1}-\frac{c}{\lambda_2}|$

where:

$f_B = 140 MHz = 140\cdot 10^6 Hz$ is the beat frequency

$\lambda_1 = 1.50 cm = 0.015 m$ is the wavelength of the first generator

$\lambda_2$ is the wavelength of the second generator

We also know that the second generator emits the longer wavelength, so we already know that the term inside the module is positive. Therefore, we can now solve for $\lambda_2$ :

$f_B = c(\frac{1}{\lambda_1}-\frac{1}{\lambda_2})\\\lambda_2=(\frac{1}{\lambda_1}-\frac{f_B}{c})^{-1}=(\frac{1}{0.015}-\frac{140\cdot 10^6}{3\cdot 10^8})^{-1}=0.015106 m = 1.5106 cm$

4 0

3 years ago

EXPLAIN CONSERVATION OF MOMEMTUM WITH THE HELP OF COLLISION OF TWO OBJECT.

Vlada [557]

i hate this question.

Conservation of momentum - is when the total momentum before and after collision is equal.

Here is the formula darling,

p = p

mv = mv

See the pic for example. HmpH

6 0

3 years ago