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sesenic [268]
3 years ago
12

a 0.4 kg block rests on a desk. the coefficient of static friction is 0.2. You push the side of the block but do not have a spri

ng scale to measure the force you use. the block does not move. which statement is true about the force of static friction?
Physics
2 answers:
ladessa [460]3 years ago
7 0
Static friction is greater than Applied force
saw5 [17]3 years ago
6 0

Answer:

D. It is no smaller than 0.78N

Explanation:

The question is incomplete. These are all the options :

A. It is no larger than 0.78N

B. It is 0.78N

C. It is 0.08N

D. It is no smaller than 0.78N

To solve this problem, we have the data of :

m=0.4kg

Where ''m'' is the mass of the block

μ = 0.2

Where ''μ'' is the coefficient of static friction

If we want to find the magnitude of the force of static friction we need to use the following equation :

F_{sf}= μ.F_{N} (I)

Where ''F_{N}'' is the normal force that the desk exerts on the block. Its magnitude is equal to the weight (because we suppose that the block rests horizontally on the desk).

The weight ''W'' can be calculated as :

W=m.g

Where ''m'' is the mass and ''g'' is the acceleration due to gravity.

The value of ''g'' is

g=9.81\frac{m}{s^{2}}

The weight of the block is

W=(0.4kg).(9.81\frac{m}{s^{2}})

W=3.924N

Now, the weight is equal to the normal force ⇒

W=F_{N}=3.924N

Using the equation (I) :

F_{sf}=(0.2).(3.924N)

F_{sf}=0.7848N

The correct option is D. It is no smaller than 0.78N

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Suppose you give a 10 Newton push to Ryan on skis (he weighs 50 kg), how much will he accelerate?
Talja [164]

Well we can just use F=ma. The force is 10N, the mass is 50 kg, solve for a. Well since we kg and N, no conversion is necessary. So just plugging in the numbers, we get

10N = 50 kg · a

\frac{10N}{50kg}=a

A newton is just \frac{kg·m}{s^{2}}

a=\frac{\frac{10kg·m}{s^{2}}}{50kg}

The s^2 and 50 kg you multiply

a=\frac{10kg·m}{50kg·s^{2}}

The kg's cancel and 10/50 is 1/5

\frac{1}{5}·\frac{m}{s^{2}}

So the acceleration is 1/5 m/s^2


3 0
3 years ago
Which physical phenomena is responsible for the earth’s sky appearing blue? scattering reflection dispersion refraction
klio [65]

Answer:  Scattering reflection

Sunlight reaches earth's atmosphere and is scattered in all directions by all the gasses and particles in the air. Blue light is seen more than others because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.

Explanation:

7 0
3 years ago
Read 2 more answers
A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,
fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

p=mv

where

m is the mass of the train

v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

m'=2m

therefore, the new momentum is

p'=m'v=(2m)v=2(mv)=2p

so, the momentum has also doubled.

7 0
3 years ago
Which wave has a greater frequency
larisa86 [58]

Answer:

A I think

Explanation:

because what is the most frequency a because it has more frequency I think I'm not that sure

5 0
3 years ago
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A charged particle moving along the x-axis enters a uniform magnetic field pointing along the z-axis. Because of an electric fie
Furkat [3]

Answer:

Explanation:

Let the charge particle have charge equal to +q .

force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along   - y axis. ( negative of y axis )

force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.

F = q ( v i x B k )  , ( velocity is along x direction and magnetic field is along z axis. )

= (Bqv) - j  

= - Bqv j

The force will be along - ve y - direction .

If we take charge as negative or - q

force due to electric field will be along - y axis .

magnetic force = F = -q ( v i x B k )

= +  Bqv j

magnetic force will be along  + y axis

So it is difficult to find out the nature of charge on the  particle from this experiment.

5 0
3 years ago
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