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ser-zykov [4K]
3 years ago
13

A toy car with an initial velocity of 5 m/s slows to a stop with an acceleration of -1.5 m/s^2.

Physics
1 answer:
enot [183]3 years ago
4 0

Car's Initial velocity (u) = 5 m/s

Car's Final velocity (v) = 0 m/s (Stop)

Acceleration of the car (a) = -1.5 m/s²

Equation used to solve this problem:

\boxed{ \bf{ {v}^{2}  =  {u}^{2}  + 2as}}

By substituting values in the equation, we get:

\rm \longrightarrow  {0}^{2}  =  {5}^{2}  +2  \times ( - 1.5) \times s \\  \\   \rm \longrightarrow 0 = 25 - 3s \\  \\  \rm \longrightarrow 25 - 3s = 0 \\  \\  \rm \longrightarrow 25 - 25 - 3s =0  - 25 \\  \\  \rm \longrightarrow  - 3s =  - 25 \\  \\  \rm \longrightarrow 3s = 25 \\  \\  \rm \longrightarrow  \dfrac{3s}{3}  =  \dfrac{25}{3}  \\  \\  \rm \longrightarrow s = 8.3 \: m

\therefore Displacement of the toy car (s) = 8.3 m

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Which two statements about an electric motor are true?
allochka39001 [22]

Options (b) and (d) are correct about an electric motor.

An electric motor is a device which coverts electrical energy into mechanical energy.It works on the principle that when a current carrying coil is placed in a magnetic field, it experiences torque. Because of that torque, the coil rotates and thus the electrical energy gets converted into mechanical (motion) energy.

7 0
3 years ago
Read 2 more answers
If a truck loses 5610 J of energy as it slows down due to an external force of 425 N, what distance will it have moved after the
lara31 [8.8K]

Answer:

13.2m

Explanation:

Step one:

given data

Energy= 5610J

Force F= 425N

Required

The distance traveled

Step two:

We know that work done is given as

WD= force* distance

so

5610=425*d

divide both sides by 425

d= 5610/425

d=13.2m

3 0
3 years ago
A 30.0-μF capacitor is connected to a 49.0-Ω resistor and a generator whose rms output is 30.0 V at 60.0 Hz. (a) Find the rms
Natali5045456 [20]

Explanation:

Given that,

Capacitor = 30μC

Resistor = 49.0Ω

Voltage = 30.0 V

Frequency = 60.0 Hz

We need to calculate the impedance

Using formula of impedance

Z=\sqrt{R^2+X_{c}^2}.....(I)

We need to calculate the value of X_{c}

Using formula of X_{c}

X_{c}=\dfrac{1}{2\pi f c}

X_{c}=\dfrac{1}{2\times\pi\times60.0\times30\times10^{-6}}

X_{c}=88.42\ \Omega

Put the value of X_{c} into the formula of impedance

Z=\sqrt{(49.0)^2+(88.42)^2}

Z=101.08\ \Omega

(a). We need to calculate the rms current in the circuit

Using formula of rms current

I_{rms}=\dfrac{V}{Z}

I_{rms}=\dfrac{30.0}{101.08}

I_{rms}=0.30\ A

The rms current in the circuit is 0.30 A.

(b). We need to calculate the rms voltage drop across the resistor

Using formula of rms voltage

V_{rms}=I_{rms}\times R

Put the value into the formula

V_{rms}=0.30\times49.0

V_{rms}=14.7\ V

The rms voltage drop across the resistor is 14.7 V

(c). We need to calculate the rms voltage drop across the capacitor

Using formula of rms voltage

V_{rms}=I_{rms}\times X_{C}

V_{rms}=0.30\times88.42

V_{rms}=26.53\ V

The rms voltage drop across the capacitor is 26.53 V.

Hence, This is the required solution.

4 0
3 years ago
Please help fast!!
daser333 [38]

The answer is C..........      

4 0
3 years ago
How much work does this force do as the particle moves along the x-axis from x = 0 to x = l? express your answer in terms of the
nydimaria [60]
<h3><u>Answer</u>;</h3>

= F0 L ( 1 - 1/e )

<h3><u>Explanation;</u></h3>

Work done is given as the product of force and distance.

In this case;

Work done  = ∫︎ F(x) dx  

                    = F0 ∫︎ e^(-x/L) dx  

                    = F0 [ -L e^(-x/L) ] between 0 and L  

                    = F0 L ( 1 - 1/e )

3 0
3 years ago
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