How much work can be done by a 20 W motor in 5 seconds?

A sportscar has a mass of 1500 kg and accelerates at 5 meters per second squared. What is the magnitude of the force acting on t

Hatshy [7]

Answer:

7500 Newtons

Explanation:

Mass of the sportscar= 1500 kg

Acceleration of the sportscar= 5m/s^2

Hence, let the Force acting on it be F

$We\ know\ that,\\Force=Mass*Acceleration\\F=ma\\\\Here,\\F=1500*5\\=7500 kg m/s^2\ or\ 7500\ Newtons$

4 0

3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

4 years ago

A tennis ball, 0.314kg, is accelerated at a rate of 164m/s2 when hit by a professional tennis player. What force does the player

Colt1911 [192]

Newton's 2nd law of motion:

Force = (mass) x (acceleration)

= (0.314 kg) x (164 m/s²)

= 51.5 newtons

(about 11.6 pounds).

Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.

~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )

5 0

3 years ago

Rafe is placing his car on a lift to change the oil and oil filter. He needs to drive up the lift 1.5 meters and the lift raises

jeyben [28]

Answer:

104 IMA

Explanation:

7 0

3 years ago

A 3250-kg aircraft takes 12.5 min to achieve its cruising

scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

$n = \frac{P_0}{P}$

$P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}$

kinetic energy $= \frac{1}{2} mv^2 =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2$

kinetic energy $= 90590389.66 kg m^2/s^2$

gravitational energy $= mgh = 3250*9.8*10000 = 315500000.00 kg m^2/s^2$

total energy $= 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2$

$P_o =\frac{409091242.28}{750} = 545454.99 j/s$

$effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49$

effeciency n = = 49%

8 0

3 years ago