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3 years ago

4. What is the acceleration of a rock if the net

Physics

1 answer:

HACTEHA [7]3 years ago

8 0

Answer:

0.75 m/s²

Explanation:

Newton's second law:

∑F = ma

3 N = (4 kg) a

a = 0.75 m/s²

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A block of aluminum at a temperature of T1 = 32 degrees C has a mass of m1 = 12 kg. It is brought into contact with another bloc

Mariulka [41]

Answer: $T=21.33 ^{\circ}$

Explanation:

Given

mass of First Block $m_1=12 kg$

Temperature $T_1=32^{\circ}$

mass of second block $m_2=0.5 m_1=6 kg$

Temperature $T_2=9^{\circ}$

Heat capacity of aluminium c=899 J/kg-K

Final Temperature acquired by both blocks at steady state

Heat loss first block =Heat gain by second block

$m_1\times c\times (32-T)=m_2\times c\times (T-9)$

$12\times 899\times (32-T)=6\times 899\times (T-9)$

$2\times 32=3T$

$T=\frac{64}{3}=21.33 ^{\circ}$

5 0

3 years ago

3. Take sugar, oil, corn syrup, a glass and water. Pour the water in the glass and then add each of the above the substances one

hoa [83]

Here are the observations

Sugar:-

Sugar is soluble in water
so It will dissolve in water .

Corn syrup:-

Corn syrup is also basically a sugar.
It will dissolve in water too .
If we shake the mixture in glass then corn syrup will be dissolved.

Oil:-

Oil is not soluble in water
Hence it won't dissolve in water.
It will float over water and make two layers

7 0

2 years ago

2. A company hires a security firm to patrol their stores and watch out for thieves. What is this an example of?

Lunna [17]

Answer:

someone hiring someone

Explanation:

4 0

3 years ago

3. Calculate the wavelength of wave that has a frequency of 4.75 x 1012Hz.

Klio2033 [76]

Frequency=v=4.75×10^12Hz
Wavelength=?

We know

$\boxed{\sf \lambda=\dfrac{C}{V}}$

$\\ \sf\longmapsto \lambda=\dfrac{3\times 10^8ms^{-1}}{4.75\times 10^{12}s^{-1}}$

$\\ \sf\longmapsto \lambda=0.631\times 10^{-4}m$

$\\ \sf\longmapsto \lambda=6.31\times 10^{-5}m$

4 0

2 years ago

An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic f

soldi70 [24.7K]

Answer:

Explanation:

To solve the exercise it is necessary to apply the concepts related to the Magnetic Field described by Faraday.

The magnetic field is given by the equation:

$B = \frac{\mu_0 I}{2\pi d}$

Where,

$\mu =$ Permeability constant

d = diameter

I = Current

For the given problem we have a change in the diameter, twice that of the initial experiment, therefore we define that:

$B_1 = \frac{\mu_0 I}{2\pi d}$

$B_2 = \frac{\mu_0 I}{2\pi 2d}$

The ratio of change between the two is given by:

$\frac{B_2}{B_1} = \frac{\frac{\mu_0 I}{2\pi d}}{\frac{\mu_0 I}{2\pi 2d}}$

$\frac{B_2}{B_1} = \frac{d}{2d}$

$\frac{B_2}{B_1} = \frac{1}{2}$

$B_2 = B_1 \frac{1}{2}$

Therefore the correct answer is D.

4 0

2 years ago