All categories

3 years ago

Draw the circuit Picture of questions below Please please help

Physics

1 answer:

kaheart [24]3 years ago

6 0

A). The diagram of the circuit is attached.

b). The resistance of the resistor, in unit of ' ohms ', is

(Voltmeter reading in volts) / (Ammeter reading in amperes) .

You might be interested in

For an electron, magnitude of force on it is Select one • Bev bev Be BIL

maw [93]

Answer:

F = Bev

Explanation:

B is magnetic field density

e is the electron charge

v is the electron velocity

6 0

3 years ago

A 4.8-kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t=0 s, the blo

garri49 [273]

Answer:

E) 80 N/m

Explanation:

Given;

mass of the block, m = 4.8 kg

displacement of the block, x = -0.5 m

velocity of the block, v = -0.8 m/s

acceleration of the block, a = 8.3 m/s²

From Newton's second law of motion;

F = ma

Also, from Hook's law;

F = -Kx

where;

k is the force constant

Thus, ma = -kx

k = -ma/x

k = -(4.8 x 8.3) / (-0.5)

k = 79.7 N/m

k ≅ 80 N/m

Therefore, the force constant of the spring is closest to 80 N/m

4 0

3 years ago

The centroid of the plane region shown is at c. use the method of composite area to determine the radius (R)? please someone hel

GenaCL600 [577]

I don’t understand what are you trying to do ?

6 0

3 years ago

A car traveling at 25m/sec, stops in 0.103 km. what is the deceleration provided by the brakes?

lozanna [386]

Answer:

-3.03 m/s²

Explanation:

v² = v₀² + 2a(x − x₀)

where v is the final velocity,

v₀ is the initial velocity,

a is the acceleration,

x is the final position,

and x₀ is the initial position.

Given:

v = 0 m/s

v₀ = 25 m/s

x = 103 m

x₀ = 0 m

Find: a

v² = v₀² + 2a(x − x₀)

(0 m/s)² = (25 m/s)² + 2a (103 m − 0 m)

a = -3.03 m/s²

Round as needed.

7 0

3 years ago

A 50-gram ball is released from rest 80 m above the surface of the Earth. During the fall to the Earth, the total thermal energy

Nady [450]

Answer:

Speed of ball just before it hit the surface is 31.62 m/s .

Explanation:

Given :

Mass of ball , m = 50 g = 0.05 kg .

Height from which it falls , h = 80 m .

Thermal energy , E = 15 J .

Now , Initial energy of the system is :

$E_i=\dfrac{mv^2}{2}+mgh$

Here , initial velocity is zero .

Therefore , $E_i=mgh=40\ J$

Now , final energy of the system :

$E_f=\dfrac{mv^2}{2}+mg(0)+15\\\\E_f=\dfrac{0.05\times v^2}{2}+15$

Since , no external force is applied .

Therefore , total energy of the system will be constant .

By conservation of energy :

$E_i=E_f\\40=\dfrac{0.05v^2}{2}+15\\\\25=\dfrac{0.05v^2}{2}\\\\v=31.62\ m/s$

Therefore , speed of ball just before it hit the surface is 31.62 m/s .

3 0

3 years ago