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2 years ago

How much time is needed to produce 720 joules of work if 90 watts of power is used

Physics

1 answer:

Sladkaya [172]2 years ago

4 0

Answer:

8 seconds

Explanation:

power (P) is defined as the rate at which work is done.

power is measured in Watts (W) , when the work done is measured in Joules (J) and time in seconds

by the definition of power,

$power=\frac{work.done}{time}\\ time=\frac{work.done}{power}\\ =\frac{720J}{90W} \\=8 seconds$

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While Bob is demonstrating the gravitational force on falling objects to his class, he drops an 1.0 lb bag of feathers from the

____ [38]

As per the question Bob drops the bag full with feathers from the top of the building.

The mass of the bag(m)= 1.0 lb

Let the air resistance is neglected.As the bag is under free fall ,hence the only force that acts on the bag is the force of gravity which is in vertical downward direction.

Here the acceleration produced on bag due to the free fall will be nothing else except the acceleration due to gravity i.e g =9.8 m/s^2

Here we are asked to calculate the distance travelled by the bag at the instant 1.5 s

Hence time t= 1.5 s

From equation of kinematics we know that -

S=ut + 0.5at^2 [ here S is the distance travelled]

For motion under free fall initial velocity (u)=0.

Hence S= 0×1.5+{0.5×(-9.8)×(1.5)^2}

⇒ -S =0-11.025 m

⇒ S= 11.025 m

=11 m

Here the negative sign is taken only due to the vertical downward motion of the body .we may take is positive depending on our frame of reference .

Hence the correct option is B.

3 0

2 years ago

Read 2 more answers

(a) What is the sum of the following four vectors in unit-vector notation? For that sum, what are (b) the magnitude, (c) the ang

avanturin [10]

6m at 0.9 radian means 6m at $51.57^0$

Since positive angles are counter clockwise

6m at $51.57^0$ can be written as 6*cos $51.57^0$ i+6*sin $51.57^0$ j = 3.73i+4.70j

5m at $-75^0$ can be written as 5*cos $(-75)^0$ i+5*sin $(-75)^0$ j = 1.294i-4.83j

4m at 1.2 radian means 4m at $68.75^0$

4m at $68.75^0$ can be written as 4*cos $68.75^0$ i+4*sin $68.75^0$ j = 1.45i+3.73j

6m at $-210^0$ can be written as 6*cos $(-210)^0$ i+6*sin $(-210)^0$ j = -5.20i-3j

a) So sum of the vector = 1.274i+0.6j

b) Magnitude = $\sqrt{1.274^2+0.6^2} = 1.98 m$

c) Angle, $tan\theta = \frac{0.6}{1.274} \\ \\ \theta = 25.22^0$

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2 years ago

Which statements describe the applications of nuclear medicine scans? Check all that apply.

coldgirl [10]

Answer:

- asses disease progression and tissue function

- utilize a biologically active molecule

- utilize a radionuclide tracer

Explanation:

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2 years ago

Read 2 more answers

A proton is confined in an infinite square well of width 10 fm. (The nuclear potential that binds protons and neutrons in the nu

kvasek [131]

Answer:

First Question

$E = 1.065*10^{-12} \ J$

Second Question

The wavelength is for an X-ray

Explanation:

From the question we are told that

The width of the wall is $w = 10\ fm = 10*10^{-15 }\ m$

The first excited state is $n_1 = 2$

The ground state is $n_0 = 1$

Gnerally the energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as

$E = \frac{h^2 }{ 8 * m * l^2 [ n_1^2 - n_0 ^2 ] }$

Here h is the Planck's constant with value $h = 6.62607015 * 10^{-34} J \cdot s$

m is the mass of proton with value $m = 1.67 * 10^{-27} \ kg$

$E = \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27}) * (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }$

=> $E = 1.065*10^{-12} \ J$

Generally the energy of the photon emitted is also mathematically represented as

$E = \frac{h * c }{ \lambda }$

=> $\lambda = \frac{h * c }{E }$

=> $\lambda = \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 1.065 *10^{-15 } }$

=> $\lambda = 1.87*10^{-10} \ m$

Generally the range of wavelength of X-ray is $10^{-8} \to 1)^{-12}$

So this wavelength is for an X-ray.

8 0

2 years ago

A civil engineer wishes to redesign the curved roadway in the example What is the Maximum Speed of the Car? in such a way that a

vlabodo [156]

Answer:

24.3 degrees

Explanation:

A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:

$a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2$

Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.

Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.

So $gsin\alpha = 4.43cos\alpha$

$\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451$

$tan\alpha = 0.451$

$\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0$

3 0

2 years ago