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Vesna [10]
3 years ago
14

A box of mass 12 kg is at rest on a flat floor. The coefficient of static friction between the box and floor is 0.42. What is th

e minimum force needed to set the box in motion across the floor?
Physics
2 answers:
Trava [24]3 years ago
7 0

Answer: 50.4N

Explanation:

Frictional force (Ff) is the force that opposes the moving force acting on a body.

This frictional force (Ff) is the product of the coefficient of friction(n) and normal reaction (R)

Mathematically, Fm = Ff = nR (since both forces are equal and opposite forces along the horizontal axis.)

NB: Fm = Ff (true for static bodies only)

Also note that the weight (W) of the body will be equal to the normal reaction (R) (forces acting along the y axis on the body) i.e W = R

R = Weight = mg = 12×10 = 120N

n = 0.42

Substituting the datas given into the formula of friction force

Ff = nR = Fm

Fm = 0.42×120

Fm = 50.4N

Therefore the minimum force needed to set the box in motion across the floor is 50.4N

Oxana [17]3 years ago
5 0
The maximum static force that can be applied is equal to the normal force*the frictional force. the normal force on the box is equal to mg since the floor is flat using 9.81m/s^2 for gravity 12kg*9.81m/s^2 = 118N multiplying the normal force by the frictional force you get a 118*.42= 49.6N so overcome the force of static friction on the box a minimum of 49.6N would need to be applied.
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A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid
Setler79 [48]

Answer:

a)

- 7.04\times10^{11} J

b)

5.03\times10^{5} N

Explanation:

a)

m = mass of the asteroid = 43000 kg

v_{o} = initial speed of asteroid = 7600 m/s

v = final speed of asteroid = 5000 m/s

W = Work done by the force on asteroid

Using work-change in kinetic energy theorem

W = (0.5) m (v^{2} - v_{o}^{2})\\W = (0.5) (43000) (5000^{2} - 7600^{2})\\W = - 7.04\times10^{11} J

b)

F = magnitude of force on asteroid

d = distance traveled by asteroid while it slows down = 1.4 x 10⁶ m

Work done by the force on the asteroid to slow it down is given as

W = - F d\\- 7.04\times10^{11} = - F (1.4\times10^{6})\\F = 5.03\times10^{5} N

4 0
3 years ago
Liquids have what kind of volume and shape
Xelga [282]
A liquid has a definite volume but takes the shape of whatever object it's in.
Brainliest please! Hope this Helps!
6 0
3 years ago
You hold a barbell in a horizontal position. The barbell’s center of mass is exactly mid-way between your two hands. A friend wa
Harrizon [31]

Explanation:

First consider that each hand works as a fulcrum: a pivot point where the barbell can rotate.

Now consider only the left hand. If the center of mass of the barbell is between hands (in the middle) it is displaced respect the fulcrum, therefore the weight which is pushing the bar downwards becomes a rotational force. The same thing happens to the other hand. Now, if more weight is added to the left hand the center of mass is displaced towards the left hand and depending how much weight is added, the center of mass will change its position and therefore the torque each hand experiences changes.

If the center of mass is still between hands: The torque remains almost the same changing only the magnitudes but not the direction.

If the center of mass is on the hand: there is no torque for the left hand because there is no leaver.

If the center of mass is to the left: now the torque changes direction and both hands need to stop it in the same direction.

(see diagram below)

6 0
3 years ago
An ancient club is found that contains 100 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assu
never [62]

Answer:

The age of living tree is 11104 years.

Explanation:

Given that,

Mass of pure carbon = 100 g

Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

We need to calculate the decay rate

R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N....(I)

Where, N = number of radio active atoms

t_{\frac{1}{2}}=half life

We need to calculate the number of radio active atoms

For N_{12_{c}}

N_{12_{c}}=\dfrac{N_{A}}{M}

Where, N_{A} =Avogadro number

N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}

N_{12_{c}}=5.02\times10^{22}\ nuclie/g

For N_{c_{14}}

N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}

N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}

N_{c_{14}}=6.526\times10^{10}\ nuclei/g

Put the value in the equation (I)

R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}

R=15.0650\ decay/min g

100 g carbon will decay with rate

R=100\times15.0650=1507\ decay/min

We need to calculate the total half lives

(\dfrac{1}{2})^{n}=\dfrac{390}{1507}

2^n=\dfrac{1507}{390}

2^n=3.86

n ln 2=ln 3.86

n=\dfrac{ln 3.86}{ln 2}

n =1.948

We need to calculate the age of living tree

Using formula of age

t=n\times t_{\frac{1}{2}}

t=1.948\times5700

t=11103.6 =11104\ years

Hence, The age of living tree is 11104 years.

5 0
3 years ago
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