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DanielleElmas [232]

2 years ago

13

What would be the weight (in Newtons) of a person with a mass of 80 kg on Earth, where the acceleration due to gravity is approx

imately 9.8 m/s/s? *
please help asap

1 answer:

defon2 years ago

6 0

Answer:

ml tayo 1v1 durugin kita

Explanation:

ano takot ml 1v1

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Darren drives to school in rush hour traffic and averages 28 mph. He returns home in mid-afternoon when there is less traffic an

Gala2k [10]

Answer:

distance between school and home is 21 miles

Explanation:

given data

in rush hour speed s1 = 28 mph

less traffic speed s2 = 42 mph

time t = 1 hr 15 min = 1.25 hr

to find out

distance d

solution

we consider here distance home to school is d and t1 time to reach at school

we get here distance equation when we go home to school that is

distance = 28 × t1 .......................1

and when we go school to home distance will be

distance = 42 × ( t - t1 )

distance = 42 × ( 1.25 - t1 ) ...................2

so from equation 1 and 2

28 × t1 = 42 × ( 1.25 - t1 )

t1 = 0.75

so

from equation 1

distance = 28 × t1

distance = 28 × 0.75

distance = 21 miles

4 0

3 years ago

he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug

S_A_V [24]

Answer:

207.4 N

Explanation:

The torque $\tau$ on a body is

$\tau = r* F$ where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

$\tau = rF\sin \theta$

Where $\theta$ is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation $\tau = rF\sin \theta$

$\tau = LF\sin \theta$

Where L is the length of the wrench.

Making F the subject

$F = \frac{\tau }{{L\sin \theta }}$

Force required to pull the wrench is given as,

$F = \frac{\tau }{{L\sin \theta }}$

Substitute $47{\rm{ N}} \cdot {\rm{m}}$ for $\tau$ , 25 cm for L, and 115o for $\theta$

$\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}$

6 0

3 years ago

Which of the following statements about X-rays and radio waves is not true? Which of the following statements about X-rays and r

Lapatulllka [165]

Answer:

X-rays travel through space faster than radio waves.

Explanation:

Electromagnetic waves consist of oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave.

All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

$c=3.0\cdot 10^8 m/s$

Electromagnetic waves are classified into 7 different types, according to their wavelength/frequency. From shortest to longest wavelength (and so, from highest to lowest frequency), we have:

Gamma rays

X rays

Ultraviolet

Visible light

Infrared radiation

Microwaves

Radio waves

Now we can analyze the 4 statements:

X-rays and radio waves are both forms of light, or electromagnetic radiation --> TRUE. They are both types of electromagnetic waves.

X-rays have higher frequency than radio waves. --> TRUE, as we can see from the table above.

X-rays have shorter wavelengths than radio waves. --> TRUE, as we can see from the table above.

X-rays travel through space faster than radio waves. --> FALSE: all electromagnetic waves travel in space at the same speed, the speed of light.

8 0

3 years ago

An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration

umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

$F=qvB$ (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

$F=qvB=ma$ (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

$B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T$

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

8 0

3 years ago

A distance of 1.0 meter separates the centers of

professor190 [17]

Neither set of choices is correct.

If the distance is tripled, then the forces decrease to

1/9 Fg. and. 1/9 Fe.

Note. When the objects are charged, the gravitational force Fg can almost always be ignored, since Fe is like 10^40 greater when the quantities of mass and charge are similar.

4 0

3 years ago