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It takes 3 minutes to make toast in a 1500 watt toaster. Calculate how much work is done by the toaster.

REY [17]

Answer: 2.7 x 10^5 joules

Explanation:

Given that:

Time taken = 3 minutes

convert time in minutes to seconds

(Since 1 minute = 60 seconds

3 minutes = 3 x 60 = 180 seconds)

Power of toaster = 1500 watt

Work done by the toaster = ?

Recall that power is the rate of work done per unit time

i.e Power = work/time

work = Power x Time

Work = 1500 watt x 180 seconds

Work = 270000J

Place the result in standard form

270000J = 2.7 x 10^5J

Thus, 2.7 x 10^5 joules of work is done by the toaster.

3 0

3 years ago

The Earth's escape speed (the speed you need to get away forever) is about 40,000 kilometers per hour. Escape speed depends on t

Anvisha [2.4K]

The Moon s escape speed will be smaller than Earth's.

What is escape speed:

The minimum speed that is required for an object to free itself from the gravitational force exerted by a massive object.

The formula of escape speed is

$v = \sqrt{\frac{2GM}{R} }$

where

v is escape velocity

G is universal gravitational constant

M is mass of the body to be escaped from

r is distance from the center of the mass

we can say that,

Escape speed depends on the gravity of the object trying to hold the spacecraft from escaping.

we know that,

The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second.

since, v ∝ g

The Moon s escape speed will be smaller than Earth's.

Learn more about escape speed here:

<u>brainly.com/question/15318861</u>

#SPJ4

5 0

2 years ago

- A cannon of 2000 kg fires a shell of 10 kg at

fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

$p=0$

After the shot, the momentum is:

$p=MV+mv$

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

$0=MV+mv$

And solving for V, we find the velocity of the cannon:

$V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s$

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

$v^2-u^2=2as$

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

$a=-0.25 m/s^2$ is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

$s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m$

7 0

3 years ago

A 1.0-kg cue ball traveling at 15 m/s strikes a stationary billiard ball of mass 1.5 kg. After the collision, the cue ball remai

Igoryamba

Answer:

Other ball's velocity is 10 m/s

Explanation:

We can use conservation of momentum:

$P_{1i} +P_{2i} =P_{1i} +P_{2i}\\1 * 15 + 1.5 * 0 = 1 * 0 + 1.5 * v\\15 = 1.5 * v\\v=\frac{15}{1.5} \,\frac{m}{s} \\v=10 \,\frac{m}{s}$

8 0

3 years ago

The position of a particle moving along an x axis is given by x = 14.0t2 - 2.00t3, where x is in meters and t is in seconds. det

Mumz [18]

a) x = $14t^{2} -2t^{3}$

at t = 5s

$x = 14*5^{2} -2*5^{3} = 14*25 - 2*125 = 350 - 250 = 100 m$

b) v = $\frac{dx}{dt}$

= $28t - 6t^2$

at t = 5 s

v = $28*5-6*25 = 140 - 150 = -10 m/s$

c) a = $\frac{dv}{dt}$

= 28 - 12t

at t = 5 s

a = 28 -12*5= 28-60= -32 m/ $s^2$

d) At maximum positive coordinate velocity = 0

So, $0 = 28t - 6t^2$

$t = \frac{28}{6} = \frac{14}{3} = 4.66 s$

At t = 4.66 s

$X = 14 * 4.66^2 - 2* 4.66^3 = 202.39 m$

e) At t = 4.66 s

f) At maximum positive velocity a = 0

$0=28-12t$

$t = \frac{28}{12} = \frac{7}{3} = 2.33 s$

At t = 2.33 s

V = $28*2.33- 6*2.33^2= 32.67 m/s$

g) t = 2.33 s

h) When particle is not moving v = 0

So $0= 28t - 6 t^2$

$t = \frac{28}{6} = 4.66 seconds$

At t = 4.66 s

a = $28 - 12 * 4. 66 = -27.93m/s^2$

i) At t = 0s, X =0m

t = 5s, X = 100m

So, Displacement = 100m

Velocity = $\frac{Displacement}{Time} = \frac{100}{5} = 20m/s$

8 0

3 years ago