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3 years ago

a toy car has a 2.0 A current, and its resistance is 1.75 ohms. How much voltage does the car require

Physics

1 answer:

o-na [289]3 years ago

5 0

Answer:

the answer will be 24.40 ohms law

Explanation:

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How should you plan to excercise if you want to remain healthy past the age of 65

gayaneshka [121]

Run,karate,aerobic exercise

5 0

1 year ago

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[1] The assembly starts from rest and reaches an angular speed of 150 rev/min under the action of a 20-N force T applied to the

ExtremeBDS [4]

Answer:

t = 5.89 s

Explanation:

To calculate the time, we need the radius of the pulley and the radius of the sphere which was not given in the question.

Let us assume that the radius of the pulley ( $r_p$ ) = 0.4 m

Let the radius of the sphere (r) = 0.5 m

w = angular speed = 150 rev/min = (150 × 2π / 60) rad/s = 15.708 rad/s

Tension (T) = 20 N

mass (m) = 3 kg each

$\int\limits^0_t {Tr_p} \, dt=H_2-H_1\\( Tr_p)t=4rm(rw)\\( Tr_p)t=4r^2mw$

$t = \frac{4r^2mw}{Tr_P}$

Substituting values:

$t = \frac{4r^2mw}{Tr_P}= \frac{4*(0.5)^2*3*15.708}{20*0.4}=5.89s$

7 0

3 years ago

Read 2 more answers

At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 fi

Amanda [17]

Answer:

Its approx location is (5.18,1.9)

Explanation:

Using F( 5,2) = ( xy-1, y²-11)

= ( 5*2-¹, 2²-11)

= (9,-5)

= so at point t=1.02

(5,2)+(1.02-1)*(9,-5)

(5,2)+( 0.02)*(9,-5)

(5+0.18, 2-0.1)

= ( 5.18, 1.9)

3 0

3 years ago

Two parallel plates that are initially uncharged are separated by 1.2 mm. What charge must be transferred from one plate to the

Gnoma [55]

Answer: 55.52 *10^-6 C= 55.52 μC

Explanation: In order to solve this question we have to take into account the following expressions:

potential energy stired in a capacitor is given by:

U=Q^2/(2*C) where Q and C are the charge and capacitance of the capacitor.

then we have:

Q^2= 2*C*U=

C=εo*A/d where A and d are the area and separation of the parallel plates capacitor

Q^2=2*εo*A*U/d=2*8.85*10^-12*1.9*10^-5*11*10^3/(1.2*10^-3)=

=55.52 *10^-6C

4 0

2 years ago

A quarterback claims that he can throw the football a horizontal distance of 167 m. Furthermore, he claims that he can do this b

pychu [463]

Answer:u=42.29 m/s

Explanation:

Given

Horizontal distance=167 m

launch angle $=33.1^{\circ}$

Let u be the initial speed of ball

Range $=\frac{u^2\sin 2\theta }{g}$

$167=\frac{u^2\sin (66.2)}{9.8}$

$u^2=1788.71$

$u=\sqrt{1788.71}$

$u=42.29 m/s$

7 0

2 years ago