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alexgriva [62]
3 years ago
15

A substance made of one or more atoms of a single type is a ?

Physics
1 answer:
nadya68 [22]3 years ago
8 0
<span>An element is a substance made up of only one type of atom.</span>
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A charge of 9 pC is uniformly distributed throughout the volumebetween concentric spherical surfaces having radii of 1.6 cmand
balu736 [363]

Answer:

Electric Field = 3.369 x 10^4 N/C

Explanation:

Radius = r = (r1 + r2) / 2 = (1.6 + 3.6) /2 = 2.6 cm + 2.3 cm = 4.9 cm = 0.049 m

As we know, Electric field = E = kQ/r.r

= 8.98755 x 10^9 x 9 x 10^-9 / 0.049 x 0.049 = 33689.275 N/C

= 3.369 x 10^4 N/C

7 0
4 years ago
How much work is done if you push a 200 N box across a floor with a force of 50 N for a distance of 20m
vovikov84 [41]
<h2>Answer: 1000 J</h2>

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

It should be noted that it is a scalar magnitude, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:  

1J=(1N)(1m)=Nm  

Now, when the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

W=(F)(d) (1)

When they are not parallel, both directions form an angle, let's call it \alpha. In that case the expression to calculate the Work is:  

W=Fdcos{\alpha} (2)

For example, in order to push the 200 N box across the floor, you have to apply a force along the distance d to overcome the resistance of the weight of the box (its 200 N).  

In this case both <u>(the force and the distance in the path) are parallel</u>, so the work W performed is the product of the force exerted to push the box F=50N by the distance traveled d. as shown in equation (1).

Hence:

W=(50N)(20m)  

W=1000Nm=1000J >>>>This is the work  

8 0
3 years ago
A voltmeter was used to check the coolant and a reading of 0.2 volt with the engine off was measured. A reading of 0.8 volt was
Julli [10]

Answer:

C. Technician B

Explanation:

Excessive Galvanic activity:

To check for excessive galvanic activity, voltmeter is used to check the coolant. If the voltmeter is giving a reading greater than 0.5 V, there is excessive galvanic activity. Excessive galvanic activity is solved by flushing the coolant fluid from engine and refiling it.

Electrolysis problem:

When the system is not properly ground, the cooling system accepts stray current and the coolant becomes an electrolyte which might eat up the radiator. To test for excessive electrolysis, start the engine and turn on all electrical accessories, if the reading is more than 0.5 V, there is electrolysis problem. Ground wires and connections should be checked at this point to stop stray current.

In our case, the first reading is 0.2 V(engine turned off) which is normal and there is no excessive galvanic activity. This means that Technician A is not correct. The second reading is 0.8 V when the engine and all electrical accessories are turned on. This reading is greater than 0.5 V which means there is an electrolysis problem. This means that Technician B is correct and ground wires and connections should be inspected and repaired.

7 0
3 years ago
¿Qué ocurre como consecuencia de las corrientes de convección en la astenosfera?
Vladimir [108]

Respuesta:En la astenosfera existen lentos movimientos de convección que explican la deriva continental. Además, el basalto de la astenosfera fluye por extrusión a lo largo de las dorsales oceánicas, lo cual hace que se renueve y expanda constantemente el fondo oceánico. :D

6 0
3 years ago
A ball, which has a mass of 1.25 kg, is thrown straight up from the top of a building 225 meters tall with a velocity of 52.0 m/
Elena-2011 [213]

First we will find the speed of the ball just before it will hit the floor

so in order to find the speed of the cart we will first use energy conservation

KE_i + PE_i = KE_f + PE_f

\frac{1}{2}mv_i^2 + mgh = \frac{1}{2}mv_f^2 + 0

\frac{1}{2}(1.25)(52)^2 + 1.25(9.8)(225) = \frac{1}{2}(1.25)v_f^2

So by solving above equation we will have

v_f = 84.3 m/s

now in order to find the momentum we can use

P = mv

P = 1.25 \times 84.3

P = 105.4 kg m/s

6 0
3 years ago
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