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vaieri [72.5K]
2 years ago
14

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

ect is:________
a. 0 n.
b. 5 n.
c. 10 n.
d. 15 n.
Physics
1 answer:
Anastaziya [24]2 years ago
6 0

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

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An electron is negatively charged.
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The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1
telo118 [61]
<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life h of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: h_{B-13}=5(10)^{-10}s=0.0000000005s

beryllium-15: h_{B-15}=2(10)^{-7}s=0.0000002s

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

h_{B-15}=X.h_{B-13}

Where X is the amount we want to find:

X=\frac{h_{B-15}}{h_{B-13}}

X=\frac{2(10)^{-7}s}{5(10)^{-10}s}

Finally:

X=400

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0
3 years ago
. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T
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Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

         I = F t = Δp

in this case the final velocity is zero

        F t = 0 -m v₀

        F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

        F₁ = -p₀ / 0.1

        F₁ = - 10 p₀

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         F₂ = -p₀ / 1

let's look for the relationship of the forces

        F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

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Answer:

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Explanation:

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8 0
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