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2 years ago

If a laptop of mass 1.4 kg is lifted to a table of height 0.8 m, how much gravitational potential energy is added to the laptop?

2 answers:

4 0

Potential Energy = mgh ,

mass = 1.4 kg, g ≈ 10 m/s², h = 0.8m

Potential Energy = mgh = 1.4 * 10 * 0.8 = 11.2 N

Potential Energy added = 11.2 N

Aleks [24]2 years ago

3 0

To calculate Potential Energy you need to Multiply 1.4*.8*9.8.... 9.8 is the acceleration due to gravity an they expect you to know to add 9.8 when you calculate Potential Energy. So the correct answer is 10.976 but rounded up is 11.

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You skip north for 12 minutes to your best friend's house that is 1.5 kilometers away. What is your average velocity?

bagirrra123 [75]

Answer:

The average velocity is 7.5 km/h

Explanation:

Let's convert minutes to hours so our answer can be given in a common units of km/hour:

12 minutes = 12/60 hours = 0.2 hours

Now we estimate the average velocity calculating the distance travelled over the time it took:

1.5 / 0.2 km/h = 7.5 km/h

3 0

2 years ago

A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a

pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

$120^2=130^2+50^2-2\times130\times50\cos\alpha$

$\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}$

$\alpha=\cos^{-1}(0.3846)$

$\alpha=67.38^{\circ}$

We need to calculate the angle β

Using cosine law

$50^2=130^2+120^2-2\times130\times120\cos\beta$

$\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}$

$\beta=\cos^{-1}(0.923)$

$\beta=22.63^{\circ}$

We need to calculate the force on 130 cm side

Using formula of force

$F_{130}=ILB\sin\theta$

$F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0$

$F_{130}=0$

We need to calculate the force on 120 cm side

Using formula of force

$F_{120}=ILB\sin\beta$

$F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63$

$F_{120}=0.1385\ N$

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

$F_{50}=ILB\sin\alpha$

$F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38$

$F_{50}=0.1385\ N$

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0

2 years ago

The brain receives messages in signals called nerve impulses. Which part of the ear first generates these nerve impulses?

liberstina [14]

That's the job of the tiny "hair cells", located in the <em>inner ear.</em>

If you're a sound wave, this is how you reach the hair cells:

-- go into the big funnel of skin on the outside of the head, that thing we call the "ear"

-- go about an inch or two, down through a skinny dark tunnel inside the skull

-- at the end of the tunnel, hit a dead end, made of a wall of thin skin like a drum, called the "ear drum"; sound waves hit the ear drum and make it vibrate

-- on the other side of the ear drum, inside, is the chamber called the "middle ear". In there are the three smallest bones in the body; the ear drum touches the first one and makes it vibrate; the first one touches the second one and makes it vibrate; the second one touches the third one and makes it vibrate; then the third one touches another dead end made of thin skin.

-- the region on the other side of this wall of thin skin is the "inner ear"; it's a long skinny chamber, called the "cochlea", wound up in a spiral and filled with liquid; the walls of the cochlea are lined with millions of tiny hairs, sticking out into the liquid; the vibrations make waves in the liquid, and the waves make the tiny hairs wave back and forth; each tiny hair is the end of a nerve that goes into the brain; when that hair wiggles, it sends a nerve "message" into the brain.

-- there are two complete copies of this whole structure ... one on each side of your head.

6 0

2 years ago

Read 2 more answers

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$