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jonny [76]
2 years ago
14

How many calories would you take in if you ate the whole box of crackers in one sitting ?

Physics
1 answer:
sweet [91]2 years ago
7 0

Answer:

i'm pretty sure it's 81 calories

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A proton and an electron are moving in the +x direction in a magnetic field in the +z
Aleksandr [31]

Answer:

Explanation:

Force on a moving charge is given by the following relation

F = q ( v x B )

for proton

q = e  ,   v = vi  , B = Bk

F = e ( vi x Bk )

= Bev - j

= - Bevj

The direction of force is along negative of y axis or -y - axis.

for electron

q = - e  ,   v = vi  , B = Bk

F = - e ( vi x Bk )

= - Bev - j

=  Bevj

The direction of force is along positive  of y axis or + y - axis.

5 0
3 years ago
This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi
Rufina [12.5K]

Answer:

 N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

            N - w₁ -w₂ =

            N = m₁ g + m₂ g

            N = g (m₁ + m₂)

let's calculate

            N = 9.8 (0.760 + 1.630)

            N = 23.4 N

This is the force of the support of the two blocks on the surface.

7 0
2 years ago
A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant bef
inessss [21]

Answer:

29.396988 m/s

Explanation:

Really, it depends on where the child is when he drops the ball - e.g., which planet he is on, and his distance from the center of that planet.

I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.

The acceleration, g, is found from

g = GM/r²

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

M = 5.9724e+24 kg

r = 6.378e+6 m

g = 9.799086 m sec⁻²

An approximate answer is found from an equation from constant acceleration kinematics:

v = gt

t = 3.0 sec

v = 29.397259 m/s

Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:

v = √[GM(2/r − 1/a)]

Here,

a = the semimajor axis of a plunge orbit, which is equal to half of the apoapsis distance of 6378000+h, where

h = the altitude from which the ball is dropped

We can (using some math) develop the following equation:

t − t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

t − t₀ = 3 sec

r = 6378000 meters

d = r + h

Using an iterative method (e.g. Newton's or Danby's), we can determine that the altitude,

h = 44.0954 meters

So,

d = 6378044.09538 meters

a = d/2 = 3189022.04769 meters

Now we can calculate that

v = 29.396988 m/s

This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.

5 0
2 years ago
Is it possible to accelerate while moving constant speed
marusya05 [52]

Answer:

Explanation:

An object which experiences either a change in the magnitude or the direction of the velocity vector can be said to be accelerating. This explains why an object moving in a circle at constant speed can be said to accelerate - the direction of the velocity changes.

3 0
3 years ago
Please solce thiss quesstion​
noname [10]

Answer:

Can you give a clearer question

Explanation:

pls give me points i have questions

4 0
3 years ago
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