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velikii [3]
3 years ago
5

Why does it hurt your hands when you carry heavy shopping in carrier bags?

Physics
2 answers:
Elodia [21]3 years ago
8 0
I think it is when the plastic is wrap tight in your hand it makes them fell sore
PilotLPTM [1.2K]3 years ago
8 0
It hurts your hands because the handle of the carrier bag cuts into your skin
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When you don't have enough room to stop, you may _______ to avoid what's in front of you.
xxMikexx [17]
Do you have the options? I would say swerve?

7 0
3 years ago
Read 2 more answers
An action force is 40 N to the right. The reaction force must be: A. 20 N left B. 40 N left C. 20 N right D. 40 N right
Sergeeva-Olga [200]

Answer:

The answer to your question is letter B.

Explanation:

To answer this question, we must remember the third law of motion of Newton that states that For every action, there is an equal and opposite reaction.

Then, if the action force is 40 N to the right, the reaction force must be 40 N to the left.

8 0
3 years ago
Archimedes' principle can be used to calculate the density of a fluid as well as that of a solid. suppose a chunk of iron with a
ella [17]
<span>(a) 39.5 g (b) 49.53 cm^3 (c) 0.7975 g/cm^3, liquid is an alcohol (a) This will be the difference between the weight of the iron in air and the weight submerged in fluid. So: 390.0 g - 350.5 g = 39.5 g (b) The density of iron is 7.874 g/cm^3, so the volume of the iron chunk is 390.0 g / 7.874 g/cm^3 = 49.53 cm^3 (c) The density of the fluid will be the mass of the fluid divided by the volume, so: 39.5 g / 49.53 cm^3 = 0.7975 g/cm^3 Since the density is very dependent upon the temperature and since the temperature wasn't specified, the actual substance can't be completely identified. Although some candidates are: 1. Mixture of Alcohol and water. Density ranges from 0.785 g/cm^3 to 1.000 g/cm^3. 2. Crude oil. Density 0.790 g/cm^3 3. Hydrazine. Density 0.795 g/cm^3 4. Methanol. Density 0.791 g/cm^3 5. Ocimene. Density 0.798 g/cm^3 The most likely candidate is a high concentration of an alcohol of some sort.</span>
5 0
3 years ago
A 25 pF parallel-plate capacitor with an air gap between the plates is connected to a 100 V battery. A Teflon slab is then inser
fgiga [73]

Answer:

5.3 nC

Explanation:

The initial charge stored on the capacitor is given by:

Q=CV (1)

where

C=25 pF = 25\cdot 10^{-12}F is the capacitance of the capacitor

V=100 V is the potential difference across the capacitor

Substituting numbers into the equation, we have

Q=(25\cdot 10^{-12} F)(100 V)=25\cdot 10^{-10}C=2.5 nC

When the Teflon slab is inserted between the plates, the capacitance of the capacitor is increased as follows:

C'=kC

where k=2.1 is the dielectric constant of the Teflon. Since the voltage V remains constant, this means that the new charge stored by the capacitor (1) will be

Q'=C'V=kCV=kQ

and so

Q'=(2.1)(2.5 nC)=5.3 nC

6 0
3 years ago
Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum
asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0
3 years ago
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