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Nonamiya [84]
3 years ago
9

Food deteriates more slowly in a refridgerator because

Physics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

<h3>Molecules have to collide with a minimum amount of energy, the activation energy or Ea . The lower the temperature, the less energy the molecules have, and the slower they therefore go. A fridge provides quite low temperatures, meaning fewer molecules have Ea or higher.</h3>
You might be interested in
A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t
borishaifa [10]

Answer:

Angular acceleration will be 18.84rad/sec^2

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity \omega _i=0rev/sec

And final angular velocity \omega _f=24rev/sec

Time is given as t = 8 sec

From equation of motion

We know that \omega _f=\omega _i+\alpha t

24=0+\alpha \times 8

\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2

So angular acceleration will be 18.84rad/sec^2

4 0
4 years ago
The surface of the inclined plane shown is frictionless. If F = 30 N, what is the
zhuklara [117]

Answer and Explanation:

5 0
4 years ago
What hanging mass will stretch a 3.0-m-long, 0.32 mm - diameter steel wire by 1.3 mm ? The Young's modulus of steel is 20×10^10
raketka [301]

Answer:

0.71 kg

Explanation:

L = length of the steel wire = 3.0 m

d = diameter of steel wire = 0.32 mm = 0.32 x 10⁻³ m

Area of cross-section of the steel wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (0.32 x 10⁻³)²

A = 8.04 x 10⁻⁸ m²

ΔL = change in length of the wire = 1.3 mm = 1.3 x 10⁻³ m

Y = Young's modulus of steel = 20 x 10¹⁰ Nm⁻²

m = mass hanging

F = weight of the mass hanging

Young's modulus of steel is given as

Y = \frac{FL}{A\Delta L}

20\times 10^{10} = \frac{F(3)}{(8.04\times 10^{-8})(1.3\times 10^{-3})}

F = 6.968 N

Weight of the hanging mass is given as

F = mg

6.968 = m (9.8)

m = 0.71 kg

7 0
4 years ago
If =12a andthe distance from each wire to point p is 0.12m, then what is the magnitude of the magnetic force per unit length on
vaieri [72.5K]

The magnitude of the magnetic force per unit length on the top wire is

2×10⁻⁵  N/m

<h3>How can we calculate the magnitude of the magnetic force per unit length on the top wire ?</h3>

To calculate the magnitude of the magnetic force per unit length on the top wire, we are using the formula

F= \frac{\mu_0 I_f}{2\pi d}

Here we are given,

\mu_0= magnetic permeability

= 4\pi×10⁻⁷ H m⁻¹

If= 12 A

d= distance from each wire to point.

=0.12m

Now we put the known values in the above equation, we get

F= \frac{\mu_0 I_f}{2\pi d}

Or, F = \frac{4\pi \times 10^{-7}\times  12}{2\pi \times 0.12}

Or, F= 2×10⁻⁵ N/m.

From the above calculation, we can conclude that the magnitude of the magnetic force per unit length on the top wire is 2×10⁻⁵ N/m.

Learn more about magnetic force:

brainly.com/question/2279150

#SPJ4

7 0
2 years ago
2. What is the percent composition of sulfur in H2SO4?
allochka39001 [22]

Answer:

C. 32.7%

Explanation:

% composition = ( mass S / mass H2SO4 ) × 100 = 32.08/ 98.10 × 100 = 32.7 % pls mark brainliest

6 0
3 years ago
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