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3 years ago

What happens when an object is moved against gravity, such as rolling a toy car up a

Physics

1 answer:

Anarel [89]3 years ago

7 0

Answer:

gravitational force remain constant

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A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc

Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringe $y_m = \frac{m\lambda D}{d}$

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

$y_1 = \frac{1\lambda D}{d}$

$y_1 = {706*D}{d}$

for dark fringe,we have

$y_m = {(m + 1/2)\lambda D}{d}$

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0

4 years ago

A car is driving at 30 m/s it has a mass of 1800 kg what is the cars momentum?

Ostrovityanka [42]

Answer:

Explanation:

Momentum is a concept and is defined as,

Momentum = mass × velocity

So to calculate the momentum of the car

momentum of the car = mass of the car × velocity of the car

So we get,

momentum of the car = 1800 × 30

= 54000 Ns

5 0

4 years ago

Read 2 more answers

A 740-kg boulder is raised from a quarry 119 m deep by a long uniform chain having a mass of 550 kg . This chain is of uniform s

Naddika [18.5K]

Answer:

A) the maximum acceleration the boulder can have and still get out of the quarry

B) how long does it take to be lifted out at maximum acceleration if it started from rest

Explanation:

let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.

the weight of the chain is: $w_{c} =m_{c} g$ and maximum tension is $T=2.50 m_{c} g=1.41*10^4N$

total mass and weight is :

$M =m_{c}+ m_{b} =740kg+550kg=1290 kg$

$w_{M} =1.2650*10^4N$

∑ $F_{y} =ma_{y}$

$T-M_{g} =Ma_{y}$

$a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg$

$=0.645m/s^2$

maximum acceleration

$a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0$

using $y-y_{0} =v_{oy} t+1/2(a_{y} )t^2$

to solve for t

$t=\sqrt{2(y-y_{0} )/a_{y} }$

$t=\sqrt{2(119m)/0.645m/s^2} =19.20s$

6 0

3 years ago

Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1

Tju [1.3M]

Answer:

Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

=nCv.dT

dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

5 0

3 years ago

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What scenario would most likely lead to a higher level of physical fitness among people who live in a city?

ExtremeBDS [4]

More people walk to school or work hope this helps

7 0

3 years ago