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3 years ago

Emmett is lifting a box vertically. Which forces are necessary for calculating the total force?

1 answer:

4 0

When Emmett is lifting a box vertically, the forces that must be added to calculate the total force are: the gravitational force, tension force(the force exerted by Emmett to the box and the force exerted by the box to Emmett), and air resistance force.

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HELPPPPPP!!!! Jupiter has a gravity that is 2.4 times that of Earth. A person has a mass of 60kg. What is the mass of this perso

mafiozo [28]

because mass of an object never change

but weight can change for example if it's

mass is 60kg 5he wieght will be 60kg * 9.8m/s²

=588N

3 0

3 years ago

Point charges q1 and q2 are separated by a distance of 60 cm along a horizontal axis. The magnitude of q1 is 3 times the magnitu

Alex777 [14]

d = distance between the two point charges = 60 cm = 0.60 m

r = distance of the location of point "a" where the electric field is zero from charge $q_{1}$ between the two charges.

$q_{1}$ = magnitude of charge on one charge

$q_{2}$ = magnitude of charge on other charge

$q_{1}$ = 3 $q_{2}$

$E_{1}$ = Electric field by charge $q_{1}$ at point "a"

$E_{2}$ = Electric field by charge $q_{2}$ at point "a"

Electric field by charge $q_{1}$ at point "a" is given as

$E_{1}$ = k $q_{1}$ /r²

Electric field by charge $q_{2}$ at point "a" is given as

$E_{2}$ = k $q_{2}$ /(d-r)²

For the electric field to be zero at point "a"

$E_{2}$ = $E_{1}$

k $q_{2}$ /(d-r)² = k $q_{1}$ /r²

$q_{2}$ /(d-r)² = 3 $q_{2}$ /r²

1/(0.60 - r)² = 3 /r²

r = 0.38 m

r = 38 cm

8 0

3 years ago

The height of a wave crest is called __________________________. a energy b rest position c amplitude d freqency e wavelength

Sladkaya [172]

B. rest position
hope this helps!!!

7 0

3 years ago

A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final

77julia77 [94]

Answer:

final velocity = 3504 m/s

Explanation:

Given data:

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

To find:

Final velocity: ?

Formula:

x = Vx*t + ½*ax*t²

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

Solution:

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²

ax = 19.2 m/s²

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s

similarly vertically:

y = Vy*t + ½*ay*t²

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²

ay = 258 m/s²

V'y = Vy + ay*t

= 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s

V = √(V'x² + V'y²)

= 3504 m/s

8 0

3 years ago

A coil with a magnetic moment of 1.50 A-m^2 is oriented initially with its magnetic moment antiparallel? to a uniform magnetic f

svp [43]

∆U = μB = (1.5)(0.815) = 1.2225 joules

6 0

3 years ago