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ruslelena [56]
3 years ago
8

If the force of static friction on a crate is 67 N and the weight of the create is 289 N, what is the coefficient of static fric

tion?
a
19363
b
0.23
c
4.31
d
222
Physics
1 answer:
RUDIKE [14]3 years ago
8 0

Answer:

B) μ = 0.23

Explanation:

The coefficient of friction is equal to μ = F / N where μ (mu) is the coefficient of friction, F is the friction force, and N is the normal force (the force of an object being applied onto the earth by gravity).

F, the friction force, is given as 67 N

Since the weight of the crate is 289 N, that means the normal force is 289 N.

This means that the coefficient of friction is  μ = F / N = 67 N / 289 N = 0.23183391 = 0.23

Therefore, B is the correct answer

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When Dr. Hewitt immerses an object in water the second time and catches the water that is displaced by the object, how does the
Dimas [21]

Answer:

Explanation:

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- The amount of Buoyancy Force is proportional to the fraction of Volume of object immersed in water; hence, the same amount is spilled/lost.

5 0
3 years ago
A steel casting weighing 2 kg has an initial temperature of 500°c; 25 kg of water initially at 25°c is contained in a perfectly
SCORPION-xisa [38]
<span> Plan: Use Q = m · c · ΔT three times. Hot casting cools ΔT_hot = 500°C - Tf. Cold water and steel tank heat ΔT_cold = Tf - 25°C. Set Q from hot casting cooling = Q from cold tank heating.
here
m_cast · c_steel · ΔT_hot = (m_tank · c_steel + m_water · c_water) · ΔT_cold

m_cast · c_steel · (500°C - Tf) = (m_tank · c_steel + m_water · c_water) · (Tf - 25°C)

2.5 kg · 0.50 kJ/(kg K°) · (500°C - Tf) = (5 kg· 0.50 kJ/(kg K°) + 40 kg· 4.18 kJ/(kg K°)) · (Tf - 25°C)

Solve for Tf, remember that K° = C° (i.e. for ΔT's) </span>
3 0
3 years ago
Read 2 more answers
Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int
neonofarm [45]

Answer:

The  value is  V_n  =  2.2498 \  m^3

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  V_n  =  3.6 \  L=  3.6 *10^{-3} \ m^3

   The  density of  nitrogen at gaseous form   is  \rho_n =  1.2929 \  kg/m^3  =  The dry air at sea level

   

Generally the density of nitrogen at liquid form is  

         \rho _l = 808 \  kg/m^3

And this is mathematically represented as

      \rho_l  =  \frac{m}{V_l }

=>   m  =  \rho_l  *  V_l

Now the density of  gaseous nitrogen is

       \rho_n  =  \frac{m}{V_n }

=>   m  =  \rho_n  *  V_n

Given that the mass is constant

       \rho_n  *  V_n  =   \rho_l  *  V_l

        1.2929*  V_n  =   808  *  3.6*10^{-3}

=>   V_n  =  2.2498 \  m^3

       

3 0
3 years ago
A mineral's chemical composition and crystal structure help determine it's ___________.
AnnyKZ [126]

Hiya!

The answer to your question is B.

Physical Properties.

~Hope this helps~

4 0
3 years ago
At a given time of​ day, the ratio of the height of an object to the length of its shadow is the same for all objects. If a 4​-f
katrin2010 [14]

Answer:35.2 ft

Explanation:

Given

height of stick =4 ft

shadow length =2.8 ft

Angle of elevation of sun is

tan\theta =\frac{4}{2.8}

let the height of tree be h

as \thetawill remain same thus

tan\theta =\frac{h}{24.64}

\frac{4}{2.8}=\frac{h}{24.64}

h=35.2 ft

8 0
3 years ago
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