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vladimir2022 [97]
3 years ago
11

Which would require a greater force -accelerating a 10g mass at 5ms-2 or a 20g mass at 2ms-2?

Physics
2 answers:
Trava [24]3 years ago
5 0
Force 1 = 10/1000 × 5 = 0.05 N
Force 2 = 20/1000 × 2 = 0.04 N

F1>F2

so the answer is Force 1
pav-90 [236]3 years ago
4 0
F= mass X speed

10g x 5
20x2


50n
40n
50>
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anyanavicka [17]
Because if your putting tension on something tensions obviously going to increase with more pressure and weight on it
7 0
3 years ago
A uniform crate with a mass of 22 kg must be moved up along the 15° incline without tipping. The force P is horizontal. Determin
Paul [167]

Answer:

F_x=208.25\ N

Explanation:

Given that,

Mass of a crate is 22 kg

It moved up along the 15 degrees incline without tipping.

We need to find the corresponding magnitude of force P. The force P is acting in horizontal direction.

It means that the horizontal component of force is given by :

F_x=F\cos\theta\\\\F_x=mg\cos\theta\\\\F_x=22\times 9.8\times \cos(15)\\\\F_x=208.25\ N

So, the horizontal component of force is 208.25 N.

6 0
3 years ago
15) On a cold day, you take in 4.2 L (i.e., 4.2 x 10-3 m3) of air into your lungs at a temperature of 0°C. If you hold your brea
Rudiy27

Answer:

4.8L ( i.e 4.8 x 10^-3 m3)

Explanation:

Step 1:

Data obtained from the question.

Initial volume (V1) = 4.2L

Initial temperature (T1) = 0°C

Final temperature (T2) = 37°C

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below

K = °C + 273

T1 = 0°C = 0°C + 273 = 273K

T2 = 37°C = 37°C + 273 = 310K

Step 3:

Determination of the final volume.

Since the pressure is constant,

Charles' Law equation will be applied as shown below:

V1 /T1 = V2/T2

4.2/273 = V2 /310

Cross multiply to express in linear form

273 x V2 = 4.2 x 310

Divide both side by 273

V2 = (4.2 x 310)/273

V2 = 4.8L ( i.e 4.8 x 10^-3 m3)

Therefore, the volume of the air in the lungs at that point is 4.8L ( i.e 4.8 x 10^-3 m3)

3 0
3 years ago
The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.
Otrada [13]

Answer:

a)   a = 3.06 10¹⁵ m / s , b)    F= 1.43  10⁻¹⁰ N, c)    F_total = 14.32 10⁻²⁶ N

Explanation:

This exercise will average solve using the moment relationship.

a ) let's use the relationship between momentum and momentum

          I = ∫ F dt = Δp

          F t = m v_{f} - m v₀

          F = m (v_{f} -v₀o) / t

 in the exercise indicates that the speed module is the same, but in the opposite direction

          F = m (-2v) / t

if we use Newton's second law

          F = m a

we substitute

            - 2 mv / t = m a

            a = - 2 v / t

let's calculate

            a = - 2 4.59 10²/3 10⁻¹³

            a = 3.06 10¹⁵ m / s

b)      F= m a

        F= 4.68 10⁻²⁶ 3.06 10¹⁵

        F= 1.43  10⁻¹⁰ N

c) if we hit the wall for 1015 each exerts a force F

            F_total = n F

            F_total = n m a

            F_total = 10¹⁵  4.68 10⁻²⁶ 3.06 10¹⁵

            F_total = 14.32 10⁻²⁶ N

8 0
3 years ago
A soccer ball collides with another soccer ball at rest. the total momentum of the balls
Juliette [100K]

I am pretty sure the answer is C.

4 0
3 years ago
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