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PSYCHO15rus [73]
3 years ago
10

Mr.smith and his wife were trying to move their new chair. Mr. Smith pulls with a force of 30N while Mrs.Smith pushes with a for

ce of 25N in the same direction. What is the net force?
Physics
1 answer:
luda_lava [24]3 years ago
7 0

Answer:

55N

Explanation:

30N + 25N = 55N

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1.Paper Chromatography. This method is often used in the food industry. ...

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3.Evaporation. ...

4Simple distillation. ...
Fractional distillation.
7 0
3 years ago
Explain the difference between what Isaac Newton and Louis de Broglie would have to say about the momentum of a particle that is
pishuonlain [190]

Answer:

Explained

Explanation:

Newton would resort to the classical mechanics and say that the momentum of the particle that is moving with a constant velocity will be given by: momentum = mass x velocity

this approach will highlight the particle nature and will not be relativistic.

De-Broglie will say that the momentum of the particle is related to its associated matter wave and the relation between them is given by:

p = \frac{h}{\lambda}

where \lambda = wavelength of the matter wave associated to the particle, h = planck's constant

andp = \gamma\times mv

thus, this highlights the wave nature of the particle and is also relativistic.

6 0
3 years ago
The idea that species change over time is called
qaws [65]
The answer is evolution. When a specifies evolves over time they change and adapt to their environment.
4 0
3 years ago
Read 2 more answers
An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases
andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0
3 years ago
A raft with the area A , thickness= h and the mass 600 kg, Floats in still water with 7 cm
elena55 [62]

<span>In this problem, we need to solve for Bubba’s mass. To do this, we let A be the area of the raft and set the weight of the displaced fluid with the raft alone as ρwAd1g and ρwAd2g with the person on the raft, </span>where ρw is the density of water, d1 = 7cm, and d2= 8.4 cm. Set the weight of displaced fluid equal to the weight of the floating objects to eliminate A and ρw then solve for m.

<span>ρwAd1g = Mg</span>

ρw<span>Ad2g = (M + m) g</span>

<span>d2∕d1 = (M + m)/g</span>

m = [(d2<span>∕d1)-1] M = [(8.4 cm/7.0 cm) - 1] (600 kg) =120 kg</span>

This means that Bubba’s mass is 120 kg.

7 0
3 years ago
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