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Elanso [62]
3 years ago
6

Help please !! physics kinematic equations

Physics
1 answer:
jarptica [38.1K]3 years ago
7 0

Explanation:

A or E I think cous I use to do this in yr 7

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A rod of length 35.50 cm has linear density (mass per length) given by λ = 50.0 + 23.0x where x is the distance from one end, an
hram777 [196]

Answer:

(a)20.65g

(b)0.19m

Explanation:

(a) The total mass would be it's mass per length multiplied by the total lenght

0.355(50 + 23*0.355) = 20.65 g

(b) The center of mass would be at point c where the mass on the left and on the right of c is the same

Hence the mass on the left side would be half of its total mass which is 20.65/2 = 10.32 g

c(50 + 23c) = 10.32

23c^2 + 50c - 10.32 = 0

c \approx 0.19m

7 0
3 years ago
A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha
denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

E = - \frac{N*\Delta \phi}{\Delta t}  

<u>Where:</u>

<em>N = is the number of turns = 1   </em>

<em>ΔΦ = ΔB*A                                            </em>

<em>Δt = is the time = 0.3 s   </em>

<em>A = is the loop of wire area = πr² = πd²/4 </em>

<em>ΔB: is the magnetic field = (0 - 1.04) T                     </em>

Hence the average induced emf is:

E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V                      

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0
4 years ago
A table exerts a 4.0 Newton force on a book which lies at rest on its top. The force exerted by the book on the table is
SOVA2 [1]
I believe the correct answer from the choices listed above is the third option. <span>The force exerted by the book on the table is equal to the force exerted by the table which is 4.0 N. The book does not move so it must be that the forces are balanced. Hope this answers the question.</span>
4 0
3 years ago
Read 2 more answers
A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

K=\frac{1}{2}mv^2

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

U=mgh

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

mg = ma

which means that the acceleration is

a= g = -9.8 m/s^2

and the negative sign means it points downward.

7 0
3 years ago
Can charging by friction only occur in solids
Sonja [21]
Yes you are right cuz charging by friction cant be done in fluids( liquid and gas)
6 0
3 years ago
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