How much force is required to move a sled 5 meters if a person uses 60 J of work?

A transformer has a secondary voltage of 140 volts and a secondary current of 3.5 amps. if the primary current is 10 amps, what

Lynna [10]

For an ideal transformer power loss is assumed to be zero

i.e. the power in primary coil due to input voltage must be equal to power in secondary coil due to output voltage

this can be written in form of equation

$V_1 i_1 = V_2 i_2$

here we know that

$V_2 = 140 volts$

$i_2 = 3.5 A$

$i_1 = 10 A{/tex]now we will use above equation[tex]140*3.5 = 10 * V_1$

$V_1 = 49 volts$

So primary coil voltage is 49 Volts

7 0

3 years ago

How to solve question #2?

Vera_Pavlovna [14]

2a for example the first one,2sec. You know that every second it moves 3metres further. So 2x3=6 but you start at 0.50m so 6+0.50=6.5

5 0

3 years ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

3 years ago

A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread non uniformly throu

Aloiza [94]

In other words a infinitesimal segment dV caries the charge
dQ = ρ dV 

Let dV be a spherical shell between between r and (r + dr): 
dV = (4π/3)·( (r + dr)² - r³ ) 
= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) 
= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) 
drop higher order terms 
= 4·π·r²·dr 

To get total charge integrate over the whole volume of your object, i.e. 
from ri to ra: 
Q = ∫ dQ = ∫ ρ dV 
= ∫ri→ra { (b/r)·4·π·r² } dr 
= ∫ri→ra { 4·π·b·r } dr 
= 2·π·b·( ra² - ri² ) 

With given parameters: 
Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) 
= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) 
= 3.77×10⁻⁸C 
= 37.7nC

6 0

3 years ago

A boat cruises 60 meters west in 10 seconds. What is its instantaneous velocity?

kari74 [83]

A - 6 meters a second ( 6m/s)

3 0

3 years ago

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