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tresset_1 [31]
3 years ago
8

The mass of an object with 500j of kenetic energy moving with a velocity of 5 m/s is

Physics
1 answer:
saw5 [17]3 years ago
7 0
The moving energy is 5,000 m/s
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An object is pulled northward with a force of 10 n and southward with a force of 15 n. the magnitude of the net force on the obj
Anastaziya [24]

The correct option is (b) 5n

As a result, there is a net downward force of 5N operating on the object.

The resultant force is the force that results from adding the vector sums of all the forces operating on an item. The combined action of all the acting forces on the object produces the same effect as the resulting force. When determining the resulting force, the direction of the forces must be taken into account.

Given;

The northward force is Fn = 10N

The southward force is Fs = 15N

Required;

The net force on the mobile phone is Fnet = ?N

The object's weight exerts downward pressure, and upward resistance exerts upward pressure. The vector sum of these two forces will be the net force.

Fnet = Fs - Fn (Considering the direction downward as positive)

Fnet= 15N - 10N

Fnet = 5N

As a result, there is a net downward force of 5 N operating on the object.

Learn more about the Force with the help of the given link:

brainly.com/question/7362815

#SPJ4

6 0
1 year ago
How can a small spark start a huge explosion? using electric forces and molecules
OLga [1]

Answer:

Small sparks can lead to huge explosion if they are left unattended.

Explanation:

Small sparks are not harmful but if these sparks happen near some hazardous material or object then it could lead to heavy explosion. If there is some chemical substance near the spark or there are magnetic lines which can explode the spark then these minor sparks could result in heavy disastrous explosion.

4 0
3 years ago
what is the energy (in j) of a photon required to excite an electron from n = 2 to n = 8 in a he⁺ ion? submit an answer to three
grin007 [14]

Answer:

Approximately 5.11 \times 10^{-19}\; {\rm J}.

Explanation:

Since the result needs to be accurate to three significant figures, keep at least four significant figures in the calculations.

Look up the Rydberg constant for hydrogen: R_{\text{H}} \approx 1.0968\times 10^{7}\; {\rm m^{-1}.

Look up the speed of light in vacuum: c \approx 2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}.

Look up Planck's constant: h \approx 6.6261 \times 10^{-34}\; {\rm J \cdot s}.

Apply the Rydberg formula to find the wavelength \lambda (in vacuum) of the photon in question:

\begin{aligned}\frac{1}{\lambda} &= R_{\text{H}} \, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\end{aligned}.

The frequency of that photon would be:

\begin{aligned}f &= \frac{c}{\lambda}\end{aligned}.

Combine this expression with the Rydberg formula to find the frequency of this photon:

\begin{aligned}f &= \frac{c}{\lambda} \\ &= c\, \left(\frac{1}{\lambda}\right) \\ &= c\, \left(R_{\text{H}}\, \left(\frac{1}{{n_{1}}^{2}} - \frac{1}{{n_{2}}^{2}}\right)\right) \\ &\approx (2.9979 \times 10^{8}\; {\rm m \cdot s^{-1}}) \\ &\quad \times (1.0968 \times 10^{7}\; {\rm m^{-1}}) \times \left(\frac{1}{2^{2}} - \frac{1}{8^{2}}\right)\\ &\approx 7.7065 \times 10^{14}\; {\rm s^{-1}} \end{aligned}.

Apply the Einstein-Planck equation to find the energy of this photon:

\begin{aligned}E &= h\, f \\ &\approx (6.6261 \times 10^{-34}\; {\rm J \cdot s}) \times (7.7065 \times 10^{14}\; {\rm s^{-1}) \\ &\approx 5.11 \times 10^{-19}\; {\rm J}\end{aligned}.

(Rounded to three significant figures.)

6 0
2 years ago
the attendance qt a seminar in a year 1 was 500 in year 2 the attendance changed by 100 what was the percent change in attendanc
TiliK225 [7]
20% change from year one to year two.
5 0
3 years ago
A 615 N student standing on a scale in an elevator notices that the scale reads 645 N. From this information, the student knows
BARSIC [14]

Answer:

The elevator must be moving upward.

Explanation:

During the motion of an elevator, the weight of the person deviates from his or her actual weight. This temporary weight during the motion is referred to as "Apparent Weight". So, when the elevator is moving downward, the apparent weight of the person becomes less than his or her actual weight.

On the other hand, for the upward motion of the elevator, the apparent weight of the person becomes more than the actual weight of that person.

Since the apparent weight (645 N) of the student, in this case, is greater than the actual weight (615 N) of the student.

<u>Therefore, the elevator must be moving upward.</u>

8 0
3 years ago
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