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3 years ago

Will the decimal point be moving to the right or to the left? How do you know?

1 answer:

5 0

Answer: Example 2.39 move decimal to the right to get rid of the decimal same for .09 move the decimal to the left to get rid of it or if it ask how many to move to the left just move that many

Explanation:

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Go around your house and hit (or tap) ten different objects. The objects should be different shapes and made of different materi

Blababa [14]

Answer:

Explanation:

go around your house and tap random objects. For example, a sink. What noise did it make? was it loud or quiet? was it soft or hard? I hope this helps

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3 years ago

Read 2 more answers

Which factors affect the gravitational force between objects? Check all that apply.

drek231 [11]

Explanation:

distance between both objects and masses of the objects

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3 years ago

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A 150 kg safe on frictionless casters is being raised at 1.20m to the bed of a truck using planks 4m long. The force needed to p

Vesnalui [34]

The only thing you need to know in order to solve this task is that <span>plank length (which is force x), should equal the increase in potential energy, so what we have now : (mass)* g * (height).
It has to look like that: </span>
<span>F * 3.0 = 150 x 9.81 x 1.20
Then solve for F, the result should be in newtones = 588N

Do hope it makes sense.</span>

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3 years ago

A steel cable with Cross Sectional Area 3.00cm² has an elastic limit of 2.40 x 10^8pascals. Find the maximum upward acceleration

bazaltina [42]

Answer:

Stress = F / A force per unit area

A = 3.00 cm^2 = 3 E-4 m^2

F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied

F/3 = 2.4E4 N if force not to exceed limit (= f)

f = M a

a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g

3 0

3 years ago

A -3.00 nc point charge is at the origin, and a second -5.50 nc point charge is on the x-axis at x = 0.800 m. find the electric

Liula [17]

The electric field produced by a single-point charge is given by

$E(r)=k\frac{q}{r^2}$

where

k is the Coulomb's constant

q is the charge

r is the distance from the charge

To find the electric field at x=0.200 m, we need to find the electric field produced by each charge at that point, and then find their resultant.

1) The first charge is $q=-3.00 nC=-3.00 \cdot 10^{-9} C$ , and it is located at x=0, so its distance from the point x=0.200 m is

$r=0.200 m-0=0.2 m$

Therefore, the electric field is

$E_1=(8.99 \cdot 10^9 Nm^2C^{-2})\frac{(3.0 \cdot 10^{-9} C)}{(0.2 m)^2}=675 N/C$

And since the charge is negative, the direction of the field is toward the charge, so toward negative x direction.

2) The second charge is $q=-5.50 nC=-5.5 \cdot 10^{-9}C$ and it is located at x=0.800 m, so its distance from the point is

$r=0.800 m-0.200 m=0.6 m$

Therefore, the electric field is

$E_2 = (8.99 \cdot 10^9 Nm^2C^{-2})\frac{(5.5 \cdot 10^{-9} C)}{(0.6 m)^2}=137.5 N$

And since the charge is negative, the direction of the field is toward the charge, so toward positive x-direction.

3) The total electric field at x=0.200 m will be given by the difference between the two fields (because they are in opposite directions). Taking the x-positive direction as positive direction, we have

$E=E_2 -E_1 =137.5 N/C/C-675 N/C=-537.5 N/C$

and the sign tells us that the field is directed toward negative x-direction.

7 0

3 years ago