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3 years ago

Question 7 of 10

Physics

1 answer:

jok3333 [9.3K]3 years ago

4 0

Answer:

when a magnet is hanged freely in air it turns in the direction of the north and south while the magnetic north pole faces the south pole of the earth and magnetic south pole faces the north pole if the earth

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Object a has a mass of 20 g

Anarel [89]

And.. where is the rest of the question?

7 0

3 years ago

A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 90.0 km/h. At that speed the forces resisting

nata0808 [166]

Answer:

1185 N

Explanation:

From Newton’s second law of motion,

F=ma where m= mass of motorcycle, a is acceleration of the motorcycle and F=Force

Net force acting on motorcycle $F_{net}$ is given by $F_{net}=F-f$

Where F is force acting on motorcycle and f is frictional force

Substituting F-f for $F_{net}$

$F_{net}=ma$ hence ma= F- f Substituting a with 3, m with 245Kg and f with 450N as provided

245*3= F- 450

F=245*3 +450= 1185 N

6 0

3 years ago

The current in some DC circuits decays according to the function I=I0e−t/τ, where I is the current at some point in time, I0 is

almond37 [142]

Answer: 1.95

Explanation:

You should start off from the decay formula and solve for τ:

$I = I_{0}e^{\frac{t}{\tau\\ } }$

$\frac{I}{I_{0}} = e^{\frac{-t}{\tau} }$

Apply inverse logarithmic function:

$ln(\frac{0.2 A}{1.2 A} ) = \frac{-t}{\tau}$

The final form will be:

$\tau=\frac{-3.5s}{ln(\frac{0.2A}{1.2A} )}$

Inputing values for I, IO, and t:

$\tau=\frac{-3.5S}{ln(\frac{0.2 A}{1.2 A} )} = 1.95$

3 0

3 years ago

To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel

alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0

3 years ago

A race car moving along a circular track has a centripetal acceleration of 15.4 m/s? If the car has

Helen [10]

Answer:

r = 58.44 [m]

Explanation:

To solve this problem we must use the following equation that relates the centripetal acceleration with the tangential velocity and the radius of rotation.

a = v²/r

where:

a = centripetal acceleration = 15.4 [m/s²]

v = tangential speed = 30 [m/s]

r = radius or distance [m]

r = v²/a

r = 30²/15.4

r = 58.44 [m]

3 0

3 years ago