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maxonik [38]
3 years ago
12

A toaster oven uses67,500 joules of energy in 45 seconds to toast a piece of bread what is the power of the oven

Physics
1 answer:
Tcecarenko [31]3 years ago
7 0

If you simply divide joules by seconds, the answer is in units
of joules/sec (watts) which is exactly what you want.

(67,500 joules) / (45 sec) = 1,500 joules/sec = 1,500 watts

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A 13-cm-diameter cd has a mass of 25 g . part a what is the cd's moment of inertia for rotation about a perpendicular axis throu
faltersainse [42]
<span>Answer: For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2. where M is the mass of the disc and R is the radius of the disc. For the axis through the edge, use parallel axis theorem. I = I(axis through center of mass) + M(distance between the axes)^2 = 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center) = 1.5 MR^2</span>
8 0
3 years ago
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Two velcro-covered pucks slide across the ice, collide and stick to one another. Their interaction with the ice is frictionless.
balu736 [363]

Answer:

<em>1. False</em>

<em>2. True</em>

<em>3. False</em>

<em>4. True</em>

Explanation:

<u>Conservation of Momentum</u>

According to the law of conservation of linear momentum, the total momentum of the system formed by both pucks won't change regardless of their interaction if no external forces are acting on the system.

The momentum of an object of mass ma moving at speed va is

p_a=m_a.v_a

The total momentum of both pucks at the initial condition is

p_1=m_a.v_a+m_b.v_b

Both pucks are moving to the right and puck B has twice the mass of puck A (let's call it m), thus

m_a=m

m_b=2m

We are given

v_a=6\ m/s\\v_b=2\ m/s

The total initial momentum is

p_1=6m+2(2m)=10m

At the final condition, both pucks stick together, thus the total mass is 3m and the final speed is common, thus

p_2=3m.v'

Equating the initial and final momentum

10m=3m.v'

Solving for v'

v'=10/3\ m/s=3.33\ m/s

1. Compute the initial kinetic energy:

\displaystyle K_1=\frac{1}{2}mv_a^2+\frac{1}{2}2mv_b^2

\displaystyle K_1=\frac{1}{2}m\cdot 6^2+\frac{1}{2}2m\cdot 2^2

K_1=18m+4m=22m

The final kinetic energy is

\displaystyle K_2=\frac{1}{2}mv'^2+\frac{1}{2}2mv'^2

\displaystyle K_2=\frac{1}{2}m\cdot 3.33^2+\frac{1}{2}2m\cdot 3.33^2

K_2=16.63m

As seen, part of the kinetic energy is lost in the collision, thus the statement is False

2. The initial speed of puck B was 2 m/s and the final speed was 3.33 m/s, thus it increased the speed: True

3. The initial speed of puck A was 6 m/s and the final speed was 3.33 m/s, thus it decreased the speed: False

4. The momentum is conserved since that was the initial assumption to make all the calculations. True

p_1=10m

p_2=3m.v'=3m(10/3)=10m

Proven

5 0
3 years ago
How are hypotheses tested?
Mandarinka [93]

Answer:

by making observation hope it's helpful

8 0
3 years ago
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A bag of cement weighing 325 N hangs in equilibrium from three wires. Two of the wires make angles of theta1=60.0 degrees and th
Murljashka [212]
The sketch of the system is: two strings, 1 and 2, are attached to the ceiling and to a third string, 3.The third string holds the bag of cement. 

The free body diagram of the weight with the string 3, drives to the tension T3 = weihgt => T3 = 325 N

The other free body diagram is around the joint of the three strings.

In this case, you can do the horizontal forces equilibrium equation as:

T1* cos(60) - T2*cos(40) = 0

And the vertical forces equilibrium equation:

Ti sin(60) + T2 sin(40) = T3 = 325 N

Then you have two equations with two unknown variables, T1 and T2

0.5 T1 - 0.766 T2 = 0

 0.866 T1 + 0.643T2 = 325

When you solve it you get, T1 = 252.8 N and T2 = 165 N

Answer: T1 = 252.8 N, T2 = 165N, and T3 = 325N



 
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3 years ago
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stich3 [128]
The answer to this question is a the estimate the speed of light
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