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3 years ago

What is another name for the magnitude of the velocity vector

Physics

1 answer:

Yuri [45]3 years ago

4 0

Answer:

the answer is from google,...

Explanation:

Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value (magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI (metric system) as metres per second (m/s) or as the SI base unit of (m⋅s−1).

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A patient is administered 20 mg of iodine-131. How much of this isotope will remain in the body after 40 days if the half-life f

igor_vitrenko [27]

The formula for half-life is:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where A is the amount of iodine-131 initially and after 40 days, t is time, h is half-life of the isotope. Let's plug in our values to the equation:

$A_{final}=20(\frac{1}{2})^{\frac{40}{8}=0.625g$

Therefore, the patient has 0.625 grams of iodine-131 after 40 days.

4 0

3 years ago

A constant force of friction 50N is acting on a body of mass 200 Kg moving initially with a speed of 15 m/s. How long does the b

Zina [86]

Answer:

F=-50N

M=200kg

U=15m/s

F=Ma

a=F/M=-50/200=-0.25

V^2-U^2=2aS

0-(15)^2=2(-0.25)S

S=-225/-0.5=450m

V=U+at

0=15-0.25t

t=-15/-0.25=60s

Explanation:

Hope this helps, let me know if you have any questions!

Have a great day.

8 0

4 years ago

Two masses are attracted by a gravitational force of 7 N.

damaskus [11]

Answer:

The answer to your question is: F = 0.4375 N. The force will be 16 times lower than with the first conditions.

Explanation:

Data

F = 7 N

F = ? if the masses is quartered

Formula

$F = \frac{Km1m2}{r2}$

Process

Normal conditions F = Km₁m₂/r² = 7

When masses quartered F = K(m₁/4)(m₂/4)/r² = ?

F = K(m₁m₂/16)/r²

F = K(m₁m₂/16r² = 7/16 = 0.4375 N

3 0

3 years ago

Calculate the ratio of the resistance of 12.0 m of aluminum wire 2.5 mm in diameter, to 30.0 m of copper wire 1.6 mm in diameter

alukav5142 [94]

Answer: 0.258

Explanation:

The resistance $R$ of a wire is calculated by the following formula:

$R=\rho\frac{l}{s}$ (1)

Where:

$\rho$ is the resistivity of the material the wire is made of. For aluminium is $\rho_{Al}=2.65(10)^{-8}m\Omega$ and for copper is $\rho_{Cu}=1.68(10)^{-8}m\Omega$

$l$ is the length of the wire, which in the case of aluminium is $l_{Al}=12m$ , and in the case of copper is $l_{Cu}=30m$

$s$ is the transversal area of the wire. In this case is a circumference for both wires, so we will use the formula of the area of the circumference:

$s=\pi{(\frac{d}{2})}^{2}$ (2) Where $d$ is the diameter of the circumference.

For aluminium wire the diameter is $d_{Al}=2.5mm=0.0025m$ and for copper is $d_{Cu}=1.6mm=0.0016m$

So, in this problem we have two transversal areas:

For aluminium:

$s_{Al}=\pi{(\frac{d_{AL}}{2})}^{2}=\pi{(\frac{0.0025m}{2})}^{2}$

$s_{Al}=0.000004908m^{2}$ (3)

For copper:

$s_{Cu}=\pi{\frac{(d_{Cu}}{2})}^{2}=\pi{(\frac{0.0016m}{2})}^{2}$

$s_{Cu}=0.00000201m^{2}$ (4)

Now we have to calculate the resistance for each wire:

Aluminium wire:

$R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}$ (5)

$R_{Al}=0.0647\Omega$ (6) Resistance of aluminium wire

Copper wire:

$R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}$ (6)

$R_{Cu}=0.250\Omega$ (7) Resistance of copper wire

At this point we are able to calculate the ratio of the resistance of both wires:

$Ratio=\frac{R_{Al}}{R_{Cu}}$ (8)

$\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}$ (9)

Finally:

$\frac{R_{Al}}{R_{Cu}}=0.258$ This is the ratio

3 0

4 years ago

Calculate the hang time of a person who moves 3 m horizontally during a 1.25-m high jump. What is the hang time when the person

jekas [21]

The duration of time for which an object stays in air is called the hang time.

For an athlete who moves 3m horizontally during a 1.25m high jump, the hang time will be the sum of the time taken by the athlete to reach the maximum height and the time taken for the athlete to reach the ground from maximum height.

Calculate the time taken t_1 by the athlete to reach the maximum height

$t_1 = \sqrt{\frac{2h}{g}}$

$t_1 = \sqrt{\frac{2(1.25)}{9.8}}$

$t_1 = 0.5s$

The athlete takes same time to reach the ground from the maximum height, so $t_2 = 0.5s$

Calculate the hang time will be

$t =t_1+t_2$

$t = 0.5+0.5$

$t = 1s$

Therefore the hang time of the athlete when he moves a horizontal distance of 3m is 1s.

Similarly, when the athlete runs 6m horizontally, then also there will not be a change in the hang time of the athlete as the hang time is independent of the horizontal distance covered.

4 0

3 years ago