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2 years ago

A Force has a size and a(n)?

2 answers:

6 0

A force is a vector quantity. As learned in an earlier unit, a vector quantity is a quantity that has both magnitude and direction. To fully describe the force acting upon an object, you must describe both the magnitude (size or numerical value) and the direction.

Gekata [30.6K]2 years ago

4 0

Answer:

Force have both magnitude and direction since its considered to be a scalar quantity

Explanation:

____________

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(a) The stick is supported by a sharp point at the middle. On the left side, a weight of 100 g is suspended at 40 cm from the mi

Jet001 [13]

Answer:

20cm

Explanation:

Hello!

remember that the condition for a body to be at rest is that the sum of its moments and its forces be zero,

To solve this problem you must draw the free body diagram of the stick (attached image) and sum up moments at point 0 (where the sharp is located), which results in the following equation

(100g)(40cm)=x(200g)

$X=\frac{(100)(40)}{(200)} =20cm$

6 0

3 years ago

2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

2 years ago

A small glass bead has been charged to + 30.0 nC . A small metal ball bearing 2.60 cm above the bead feels a 1.80×10−2 N downwar

BlackZzzverrR [31]

Answer:

The charge on the ball bearing 4.507 × 10^-8 C

Explanation:

From Coulomb's law

F = kq1q2/r²

make q2 the subject

q2 = Fr²/kq1

q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)

q2 = 4.507 × 10^-8 C

8 0

3 years ago

Suppose the electrons and protons in 1g of hydrogen could be separated and placed on the earth and the moon, respectively. Compa

MAXImum [283]

Answer:

The gravitational force is 3.509*10^17 times larger than the electrostatic force.

Explanation:

The Newton's law of universal gravitation and Coulombs law are:

$F_{N}=G m_{1}m_{2}/r^{2}\\F_{C}=k q_{1}q_{2}/r^{2}$

Where:

G= 6.674×10^−11 N · (m/kg)2

k = 8.987×10^9 N·m2/C2

We can obtain the ratio of these forces dividing them:

$\frac{F_{N}}{F_{C}}=\frac{Gm_{1}m_{2}}{kq_{1}q_{2}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{m_{1}m_{2}}{q_{1}q_{2}}$ --- (1)

The mass of the moon is 7.347 × 10^22 kilograms

The mass of the earth is 5.972 × 10^24 kg

And q1=q2=Na*e=(6.022*10^23)*(1.6*10^-19)C=9.635*10^4 C

Replacing these values in eq1:

$\frac{F_{N}}{F_{C}}}}=0.742\times10^{-20}\frac{C^{2}}{kg^{2}}\frac{7.347\times5.972\times10^{46}kg^{2}}{(9.635\times10^{4})^{2}}$

Therefore

$\frac{F_{N}}{F_{C}}}}=3.509\times10^{17}$

This means that the gravitational force is 3.509*10^17 times larger than the electrostatic force, when comparing the earth-moon gravitational field vs 1mol electrons - 1mol protons electrostatic field

7 0

3 years ago

Two gliders are on a frictionless, level air track. Both gliders are free to move. Initially, glider A moves to the right and gl

Yuliya22 [10]

Answer:

The change in momentum of both objects is the same but in opposite direction.

Explanation:

Hi there!

The momentum of the system is calculated as the sum of the momentums of each glider. The momentum of the system is conserved if no external force is acting on the objects (as in this case). That means that the initial momentum of the system is equal to the final momentum of the system.

The momentum of each glider is calculated as follows:

p = m · v

Where:

p = momentum.

m = mass of the glider.

v = velocity.

The momentum of the system for glider A and B can be calculated as follows:

initial momentum = mA · vA + mB · vB

Where:

mA and vA = mass and velocity of glider A

mB and vB = mass and velocity of glider B

Initially, glider B is at rest so that vB = 0. Then, the initial momentum of the system is:

initial momentum = mA · vA

The final momentum of the system is calculated as follows:

final momentum = mA · vA´ + mB · vB´

Where vA´ and vB´ are the final velocities of glider A and B respectively.

We know that mB = 4mA and that vA´ is negative. The the final momentum will be:

final momentum = -mA · vA´ + 4mA · vB´

Since initial momentum = final momentum:

mA · vA = -mA · vA´ + 4mA · vB´

mA · vA + mA · vA´ = 4mA · vB´

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The change in momentum of glider A (ΔpA) is calculated as follows:

ΔpA = final momentum - initial momentum

ΔpA = -mA · vA´ - mA · vA = -mA (vA + vA´) = -4mA · vB´

The change in momentum of glider B (ΔpB) is calculated as follows:

ΔpB = final momentum - initial momentum

ΔpB = 4mA · vB´ - 0 = 4mA · vB´

Then, the change in momentum of both objects is the same but in opposite direction. That´s why the momentum is conserved.

4 0

3 years ago