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Andru [333]
2 years ago
8

consider a charge of -15.0 mCmoving to the right a 2.00x10^6 m/s in a mganetic field of .0300 T pointing upwards. What is the ma

gitude and direction of the force on this charge? If this charge had a mass of 8.00 x10^10kg what would be the acceleration be
Physics
1 answer:
77julia77 [94]2 years ago
5 0

Answer:

a = 1.13×10^-8 m/s²

Explanation:

f = ma = qvb \\ a =  \frac{qvb}{m}

q = 15×10^-3 C

v = 2×10^6 m/s

B = 0.03 T

m = 8.00 x10^10 kg

a = 900/8 x10^10

a = 1.13×10^-8 m/s²

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Answer:

density of cube =11.605 g/cm³

Explanation:

density of a substance is the mass per unit volume of that substance.

the density of a substance = \frac{mass}{volume}

volume of a cube = l³,

l = 19.0mm , lets convert mm to cm

1mm = 0.1cm, thus, 19mm =19*0.1 =1.9cm

length of cube =1.9cm

volume of cube = 1.9³

density of cube = \frac{79.6}{1.9^{3} }

density of cube =11.605 g/cm³

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3 years ago
A bulb pile is driven to the ground with a 2.5 ton hammer. The drop height is 22 ft and the volume in last batch driven is 4 cu
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Answer:

159.1 ton

Explanation:

The solution is shown in the attached file

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5 0
3 years ago
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Two groups of students were tested to compare their speed working math problems,Each group was given the same problems.One group
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Your hypothesis can be that the group with calculators would finish faster than the group without calculators.
4 0
3 years ago
How long should a spring be stretched for it to store 45 J of energy? The force constant of the spring is
ELEN [110]

Answer:

x = 0.4 m

Explanation:

When a spring is stretched from its equilibrium position. Some energy is stored in the spring. This energy is called the elastic potential energy of the spring. The formula used to calculate the magnitude of this stored energy is given as follows:

P.E = (1/2)kx²

where,

P.E = Elastic Potential Energy Stored in the spring = 45 J

k = Spring Constant = 540 N/m

x = amount of stretching = ?

Therefore,

45 J = (1/2)(540 N/m)x²

x² = (45 J)(2)/(540 N/m)

x = √(0.167 m²)

<u>x = 0.4 m</u>

5 0
3 years ago
You have covered a grounded metal surface with a layer of photoconductor. Working in the dark, you sprinkle negative charge onto
Keith_Richards [23]

Answer:

A. the left half becomes neutral while the right half remains negatively charged

Explanation:

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