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alexandr1967 [171]
3 years ago
14

When helping someone whose car has a "dead" battery, how should your car's battery be connected in relation to the dead battery?

Physics
1 answer:
leonid [27]3 years ago
8 0
Positive terminal with Positive terminal. The negative terminal should be connected to the car chassis.
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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?
stich3 [128]
I would think that you would have to do 42/2=21Hz, but I'm not sure...
5 0
3 years ago
In 2h2so4, the number of hydrogen atoms is 1 2 4 8
pishuonlain [190]
C.) In this, number of Hydrogen atoms is 4
7 0
3 years ago
An astronaut on the moon places a package on a scale and finds its weight to be 10. N. () What would the weight of the package
jeka94

Answer:

(a) 61.25 N

(b) 6.25 kg

(c) 6.25 Kg

Explanation:

Weight on moon = 10 N

Acceleration due to gravity on moon = 1.6 m/s^2

Acceleration due to gravity on earth = 9.8 m/s^2

Let m be the mass of the package.

(a) Weight on earth = mass x acceleration due to gravity on earth

Weight on earth = 6.25 x 9.8 = 61.25 N

(b) Weight on moon = mass x acceleration due to gravity on moon

10 = m x 1.6

m = 6.25 kg

(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.

8 0
3 years ago
You are driving at 40 miles per hour and you speed up to 60 miles per hour over the course of 20 miles. How long were
notka56 [123]

Answer:

120 hours

Explanation:

4 0
3 years ago
A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in
koban [17]

Answer:

a) The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) The electric force is 2.0\times 10^{-6} newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

E = \frac{F_{e}}{q} (1)

Where:

E - Electric field, measured in newtons per coulomb.

F_{e} - Electric force, measured in newtons.

q - Electric charge, measured in coulombs.

If we know that F_{e} = 4.0\times 10^{-6}\,N and q = 2.0\times 10^{-8}\,C, then the electric field at that point is:

E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}

E = 2.0\times 10^{2}\,\frac{N}{C}

The electric field at that point is 2.0\times 10^{2} newtons per coulomb.

b) If we know that E = 2.0\times 10^{2}\,\frac{N}{C} and q = 1.0\times 10^{-8}\,C, then the electric force is:

F_{e} = E\cdot q

F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)

F_{e} = 2.0\times 10^{-6}\,N

The electric force is 2.0\times 10^{-6} newtons.

7 0
3 years ago
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