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Lubov Fominskaja [6]
3 years ago
13

In a lab, a block weighing 80 N is attached to a spring scale, and both are pulled to the right on a horizontal surface. Frictio

n between the block and the surface is negligible. What is the acceleration of the block when the scale reads 32N?
Physics
2 answers:
notka56 [123]3 years ago
5 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

There should be an image that should accompanied your question, I was able to chcek it from other sources. the acceleration of the block when the scale reads 32N is <span>4.0 m/s*s</span>
Alika [10]3 years ago
4 0

Answer:

4.0 m/s 2

Explanation:

it says to find the acceleration which is change of velocity over change in time

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Describe J.J Thomson's plum pudding model,and explain how it compares to the modern day atomic theory.​
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The correct answer to this open question is the following.

The J.J Thomson's plum pudding model establishes that when he researched the atom, he found that electrons are positively charged. Because he observed that electrons were embedded like plums. Electrons are negative. However, compared to the modern-day atomic theory, scientist Ernest Rutherford established that the space between an atom is empty, the nucleus of the atom is positive, and negative electrons have fixed motion paths.

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En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

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NISA [10]

Explanation:

It is given that,

Diameter of loop, d = 1.4 cm

Radius of loop, r = 0.7 cm = 0.007 m

Magnetic field, B=2.5\ mT=2.5\times 10^{-3}\ T

(A) Magnetic field of a current loop is given by :

B=\dfrac{\mu_oI}{2r}

I is the current in the loop

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.5\times 10^{-3}\times 0.007}{4\pi \times 10^{-7}}

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(B) Magnetic field at a distance r from a wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 27.85}{2\pi \times 2.5\times 10^{-3}}

r = 0.00222 m

r=2.2\times 10^{-3}\ m

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The correct answer would be True!
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