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4 years ago

If an object has a positive acceleration that means:

Physics

1 answer:

borishaifa [10]4 years ago

7 0

Answer:

When an object is speeding up, the acceleration is in the same direction as the velocity. Thus, this object has a positive acceleration. In Example B, the object is moving in the negative direction (i.e., has a negative velocity) and is slowing down

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A 4800 Watt heater is supposed to be plugged into a 120 volt outlet to work correctly. If the heater was plugged into a 90 volt

dedylja [7]

Answer:

3600

Explanation:

90 of 120 is 3/4 so 4800 - 1200 is 3600.

3 0

3 years ago

Matter is made up of tiny particles which can be atoms,molecules, or electrically charged particles called?.

yuradex [85]

Answer: alpha particle. i think

Explanation:

6 0

3 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago

An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the os

aniked [119]

Answer:

(a) T = 0.412s

(b) f = 2.42Hz

(d) k = 86.75N/m

(e) vmax = 5.03 m/s

Explanation:

Given information:

m: mass of the block = 0.373kg

A: amplitude of oscillation = 22cm = 0.22m

T: period of oscillation = 0.412s

(a) The period is the time of one complete oscillation = 0.412s

The period is 0.412s

(b) The frequency is calculated by using the following formula:

$f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz$

The frequency is 2.42 Hz

(c) The angular frequency is:

$\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}$

The angular frequency is 15.25 rad/s

(d) The spring constant is calculated by solving the following equation for k:

$\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}$

The spring constant is 86.75N/m

(e) The maximum speed is:

$v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}$

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

$F=kA=(86.75N/m)(0.2m)=17.35N$

The maximum force that the spring exerts on the block is 17.35N

3 0

4 years ago

Give reason standard physical quantity is used for correct measurement of the given body

PtichkaEL [24]

Answer:

Physical quantities are the substances which can be measured. By measuring physical quantities, we can measure the mass, weight and other physical properties

4 0

3 years ago

Read 2 more answers