Ask question

Ask question

All categories

1 month ago

13

if a toaster transfers 100 joules of energy every ten seconds, what is the power rating of the toaster include the units in your

answer

2 answers:

Sav [3.1K]1 month ago

8 0

Answer:10 watts

Explanation:

The formula for power is work divided by time

This leads to Power=100 divided by 10

Thus, Power equals 10

Resulting in Power = 10 watts

Softa [3K]1 month ago

8 0

Answer:

this is the answer to the question

You might be interested in

A baseball thrown at an angle of 60.0° above the horizontal strikes a building 16.0 m away at a point 8.00 m above the point fro

Ostrovityanka [3204]

Answer:

a) $v_{o} =16m/s$

b) $v=9.8m/s$

c) $\beta =-35.46º$

Explanation:

According to the problem, the distance from the building where the ball hits is 16m, and its final elevation exceeds the initial height by 8m.

With this information, we can compute the ball’s starting speed.

a) Let's first assess the horizontal trajectory.

$x=v_{ox}t$

$x=v_{o}cos(60)t$

$v_{o}=\frac{x}{tcos(60)}=\frac{16m}{tcos(60)}$ (1)

This gives us our initial equation.

Next, we need to examine the vertical trajectory.

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^2$

$y_{o}+8=y_{o}+v_{o}sin(60)t-\frac{1}{2}(9.8)t^2$

Utilizing $v_{o}$ in our first equation (1)

$8=\frac{16}{tcos(60)}sin(60)t-\frac{1}{2}(9.8)t^2$

$\frac{1}{2}(9.8)t^2=16tan(60)-8$

Now let’s solve for t.

$t=\sqrt{\frac{2(16tan(60)-8)}{9.8} } =2s$

The ball takes two seconds to reach the adjacent building, allowing us to compute its initial speed.

$v_{o}=\frac{16m}{(2s)cos(60)}=16m/s$

b) To determine the velocity magnitude just before impact, we must calculate both x and y components.

$v_{x}=v_{ox}+at=16cos(60)=8m/s$

$v_{y}=v_{oy}+gt=16sin(60)-(9.8)(2)=-5.7m/s$

The computed velocity magnitude is:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{(8m/s)^2+(-5.7m/s)^2}=9.8m/s$

c) The ball's angle is:

$\beta=tan^{-1}(\frac{v_{y} }{v_{x}})=tan^{-1}(\frac{-5.7}{8})=-35.46º$

4 0

3 months ago

A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer

Keith_Richards [3271]

Response:

Clarification:

Provided

weight of disk $m=12.5 kg$

diameter of disc $R=0.23 m$

weight of ring $m_r=7 kg$

Force $F=9.7 N$

$N=180 rpm$

$\omega =\frac{2\pi N}{60}$

$\omega =6\pi rad/s$

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

$=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2$

$=13.25\times 0.23^2=0.7009 kg-m^2$

At this point, Torque is $T=F\times R=I\cdot \alpha$

$9.7\times 0.23=0.7\times \alpha$

$\alpha =3.18 rad/s^2$

Utilizing $\omega _f=\omega +\alpha t$

$\omega _f=0$ in this scenario

$0=6\pi -3.18\times t$

$t=\frac{6\pi }{3.18}$

$t=5.92 s$

7 0

2 months ago

Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th

Softa [3030]

Response:

The new resistance is half of the original resistance.

Explanation:

Resistance in a wire is represented by:

$R=\dfrac{\rho L}{A}$

$\rho$ = resistivity of the material

L and A are the physical dimensions

If a wire is exchanged for one where all linear dimensions are doubled, i.e. l' = 2l and r' = 2r

The updated resistance of the wire can be calculated as follows:

$R'=\dfrac{\rho L'}{A'}$

$R'=\dfrac{\rho (2L)}{\pi (2r)^2}$

$R'=\dfrac{1}{2}\dfrac{\rho L}{A}$

$R'=\dfrac{1}{2}R$

The new resistance equals half of the original resistance. Thus, this provides the solution needed.

4 0

2 months ago

A child on a 2.4 kg scooter at rest throws a 2.2 kg ball. The ball is given a speed of 3.1 m/s and the child and scooter move in

Maru [3345]

Answer:

The mass of the child is 14.133 kg

Explanation:

According to the principle of linear momentum conservation, we get;

(m₁ + m₂) × v₁ + m₃ × v₂ = (m₁ + m₂) × v₃ - m₃ × v₄

The negative sign signifies that their velocities were directed oppositely

Considering both the child and the ball are initially at rest, we have;

v₁ = 0 m/s and v₂= 0 m/s

Thus;

0 = m₁ × v₃ - m₂ × v₄

(m₁ + m₂) × v₃ = m₃ × v₄

Where:

m₁ = Mass of the child

m₂ = Mass of the scooter = 2.4 kg

v₃ = Final velocity of the child and scooter = 0.45 m/s

m₃ = Mass of the ball = 2.4 kg

v₄ = Final velocity of the ball = 3.1 m/s

Substituting the known values gives us;

(m₁ + 2.4) × 0.45 = 2.4 × 3.1

(m₁ + 2.4) = 16.533

This implies m₁ + 2.4 = 16.533

Thus, m₁ = 16.533 - 2.4 = 14.133 kg

The mass of the child equals 14.133 kg.

3 0

1 month ago