All categories

3 years ago

A bag of groceries has a weight if 44 newtons , what is its weight in kilograms ?

Physics

2 answers:

Oxana [17]3 years ago

3 0

Answer:

0.101971621297793 kg.

Explanation:

1 Newton: 1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth. This is derived using Newton's second law f=ma and assuming Earth gravity of 9.80665 m/s2. 1 N (Earth) = 0.101971621297793 kg.

Alchen [17]3 years ago

3 0

｡☆✼★ ━━━━━━━━━━━━━━ ☾

1 kg = 9.81 newtons.

Divide by 9.81:

44/9.81 = 4.4852...

Thus, your answer is 4.89 kg

Have A Nice Day ❤

Stay Brainly! ヅ

- Ally ✧

｡☆✼★ ━━━━━━━━━━━━━━ ☾

You might be interested in

A rock is thrown upward with a velocity of 23 meters per second from the top of a 25 meter high cliff, and it misses the cliff o

Nana76 [90]

Vf=vi+at
-23m/s=23m/s-9.8m/s^2(t)
t1 from throwing up back to starting position t1=4.693s
Delta(x)=vi(t)+.5a(t^2)
25-11=14
-14m=-23m/s(t)-4.8m/s^s(t^2)
t2 is from 25m to 11 m t2=.546s
Total time =5.239s

8 0

4 years ago

Two narrow slits 0.02 mm apart are illuminated by light from a CuAr laser (λ = 633 nm) onto a screen. a) If the first fringe is

Masja [62]

Answer:

a) 0.063m

b) 2.72°

c) 3151 fringes

d) 1.87*10^-6m

Explanation:

a) To find the screen distance you use the following formula:

$y=\frac{m\lambda D}{d}\\\\D=\frac{dy}{m\lambda}$

D: screen distance

d: distance between slits

m: order of the fringes

λ: wavelength

By replacing the values of the parameters you obtain:

$D=\frac{(0.02*10^{-3}m)(0.2*10^{-2}m)}{(1)(633*10^{-9}m)}=0.063m$

b) The condition for dark fringes is given by:

$\lambda(m+\frac{1}{2})=dsin\theta$

for the first dark fringe the angle is:

$\theta=sin^{-1}(\frac{\lambda(m+\frac{1}{2})}{d})\\\\\theta=sin^{-1}(\frac{(633*10^{-9}m)(1+\frac{1}{2})}{0.02*10^{-3}m})=2.72\°$

c) the visible number of fringes is given by:

$N=1+2\frac{D}{d}=1+\frac{0.063m}{0.02*10^{-3}}=3151 \ fringes$

d) the wavelength of a laser in which its first order fringe coincides with the third one of the CuAr laser is:

$y=\frac{(3)(633*10^{-9}m)(0.063m)}{0.02*10^{-3}m}=5.98*10^{-3}m\approx0.59cm\\\\\lambda'=\frac{dy}{mD}=\frac{(0.02*10^{-3}m)(0.59*10^{-2}m)}{(1)(0.063m)}=1.87*10^{-6}m$

4 0

3 years ago

The helicopter was deformed and destroyed in the _______ collision.

ASHA 777 [7]

There are two types of collision in physics which are named as elastic collision and inelastic collision.

Elastic collision is that type of collision in which both kinetic energy and momentum is conserved i.e total kinetic energy and momentum before collision is equal to the total kinetic energy and momentum after collision.As kinetic energy is conserved here,so less or ideally no damage occurs to the colliding bodies.

Inelastic collision is that type of collision in which only momentum is conserved but not kinetic energy .After the collision the kinetic energy is lost in forms heat,sound ,light etc.

Normally any deformation and destruction is due to the inelastic collision which results in the loss of kinetic energy.

Hence the helicopters were destroyed and deformed due to inelastic collision.

7 0

3 years ago

Read 2 more answers

How could you prove that

Lubov Fominskaja [6]

Answer:

An unconformity shows where some rock layers have been lost because of erosion. To date rock layers, geologists first give a relative age to a layer of rock at one location and then give the same age to matching layers at other locations. Certain fossils, called index fossils, help geologists match rock layers.

Explanation:

5 0

3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$