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kupik [55]
3 years ago
14

A 50kg block is at rest on the ground.

Physics
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

(b) 490 N downwards

Explanation:

(b) Formula : w= mg

m is the mass of the object

g is a gravitational constant (9.8)

w= mg

= 50×9.8

= 490 N downwards

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A magnet attracts a piece of iron. The iron can then attract another piece of iron. On the basis of domain alignment, explain wh
raketka [301]

Answer :

The magnet produces a domain alignment that allows the iron become magnetic such that it is attracted to the original magnet.the magnetic iron can then create the same effect on another piece of iron.

Explanation:

the piece of magnet is brought closer to the iron where the domains of the iron align including the magnetic poles.

5 0
4 years ago
The hottest stars are _____. -Blue -Orange -White -Red
AnnyKZ [126]

Answer:

yellow

Explanation:

6 0
4 years ago
Read 2 more answers
A man in a lift is moving upwards in a constant speed.the weight of the man is W.Acc
Nookie1986 [14]

Answer:

Normal force=mg

Explanation:

The reaction force of weight is the normal force.

in order to find the normal for we need to write all the forces and set it equal to the net force:

N-mg=ma (since it is a constant speed the a=0)

N=mg

3 0
3 years ago
Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl
tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s

For Train 2 Velocity of the Source,

v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s

Frequency of Source,

f_{s}=500\ Hz

To Find:

Frequency of Observer,

f_{o}=?  (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source v_{s} and observer v_{o}, the original frequency of the sound waves f_{s} and the observed frequency of the sound waves f_{o},

The Equation is

f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

7 0
3 years ago
PLease Help!
scoundrel [369]

Answer:

Option A is the correct answer.

Explanation:

The instantaneous acceleration = Change in velocity in velocity/Time taken

The slope of the graph should give instantaneous acceleration.

 Slope of a graph = Change in value of Y -axis / Change in values of X -axis

 Comparing both the equations

   Change in value of Y -axis = Change in velocity in velocity

   Change in values of X -axis = Time taken

So velocity values should be on the Y axis and Time values should be on the X axis.

Option A is the correct answer.

8 0
4 years ago
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