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3 years ago

7

The current in a hair dryer measures 11 amps. The resistance of the hair dryer is 15 ohms. What is the voltage? Your Answer: Que

stion 12 options: Answer: units:

1 answer:

lorasvet [3.4K]3 years ago

5 0

V=IR
V=15x11
V=165ohms

I don’t quite remember the unit

Ohms law

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¿ qué es un motor de explosión?

ArbitrLikvidat [17]

Answer:

es un motor de combustión interna con encendido por chispa.

5 0

3 years ago

A boat moves through the water with two forces acting on it. One is a 1,800-N forward push by the water on the propeller, and th

baherus [9]

(a)The acceleration of the 1,400-kg boat will be 0.425 m/sec²

(b) If it starts from rest, the distance through which the boat moves in 20.0 s will be 85 m.

(c)Velocity at the end of that time will be 8.5 m/sec.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

Given data;

Forwad force acting on the boat,v₁ = 1,800-N

Resistive force acting on the boat,v₂ = 1,200-N

The acceleration of the boat,a

Mass of boat,m = 1,400-kg

Initial velocity of boat,u= 0 m/sec

Distance travelled by boat,S = ?

Time for the boat travels,t = 20.0 s

Final velocity,V = ? m/sec

The net force on the boat;

F = F₁ - F₂

F = 1800 N - 1200 N

F = 600 N

From the defination of force;

F= ma

a = F / m

a = 600 N / 1400 kg

a = 0.425 m/sec²

b)

The distance through which the boat moves is 20.0 s;

$\rm x_f = x_ 0 + v_0 t + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}at^2 \\\\ x_f = 0+0 + \frac{1}{2} at^2 \\\\ x_f = \frac{1}{2}\times 0.425 \times (20 )^ 2 \\\\ x_f = 85 \ m$

c)

The velocity at the end of that time is found as;

$\rm v_f = v_ 0 + at \\\\ v_f =- 0+ at \\\\ v_f = 8.5$ m/sec

Hence the acceleration of the boat, the distance through which the boat moves in 20.0 s, and velocity at the end of that time will be 0.425 m/sec²,85 m, and 8.5 m/sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972

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8 0

2 years ago

The speed of an arrow fired from a compound

san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

$X=v_{ox}*t$

$v_{0x} =v_0cos(\alpha)$

Y axis:

$Y= Y_0 +v_{y0} t - \frac{g}{2} t^2$

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, $Y=0$ , then

$0= Y_0 +v_{y0} t - \frac{g}{2} t^2$

$0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

Caculate the segcond degree equation to obtain the two possible values for t:

$t_1= 10.18 \\t_2= -0.04046$

But, in physics, time it could not be negative, so we take $t_1= 10.18$

Caculate now:

$X=79m/s*cos(\39)*10.18s= 624.996 m$

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

$v_0= 79m/s+13m/s= 92m/s$

Using the same procedure that item A, caculate X

First, we need to know the new time

$0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

And we obtain:

$t_1=11.845s\\t_2=-0.041s$

One more time, we take the positive time: $t_1=11.845s$

Finally:

$X=92m/s *cos(39)*11.845s=846.887 m$

6 0

3 years ago

A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance

seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

4 0

3 years ago

Timed! I would really appreciate some help! thank you!

GenaCL600 [577]

Answer:

x = 5[km]

Explanation:

We must convert the time from minutes to hours.

$t=30[min]*\frac{1h}{60min}= 0.5[h]\\$

We know that speed is defined as the relationship between space and time.

$v=x/t$

where:

x = space [m]

t = time = 0.5 [h]

v = velocity [m/s]

Now replacing:

$x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]$

4 0

3 years ago