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mixer [17]
3 years ago
5

Bonnie and Clyde are sliding a 323 kg bank safe across the floor to their getaway car. The safe slides with a constant speed if

Clyde pushes from behind with 383 N of force while Bonnie pulls forward on a rope with 343 N of force. What is the coefficient of friction between the safe and the floor?
Physics
1 answer:
Harrizon [31]3 years ago
4 0

Answer:

835/(320*9.8)=0.266

Explanation:

since the safe is moving at a constant speed then the net force is zero

in other words the friction force equals the sum of Bonnie and Clyde force

so the friction force must be 485+350=835N

the frictional force = mu*m*g

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A race car travels 44.3 m/s around a banked (45° with the horizontal) circular (radius = 200 m) track. What is the magnitude of
djverab [1.8K]

The magnitude of the resultant force is given by the centripetal force, since the car is under a circular motion. So, we have:

F_c=ma_c

The centripetal acceleration is given by:

a_c=\frac{v^2}{r}

Where v is the linear speed and r the radius of the circular motion. Replacing this and solving:

F=m\frac{v^2}{r}\\F=80kg\frac{(44.3\frac{m}{s})^2}{200m}\\F=785N*\frac{1kN}{1000N}\\F=0.785kN

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a physical science textbook has a mass of 2.2 kg. If the textbook weighs 19.6 newtons on Venus, what is the strength of gravity
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WILL GIVE BRAINLIEST TO THE CORRECT ANSWER
LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

F = 1000*3\\F=3000[N]

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

W=m*g\\W=1000*9.81\\W=9810 [N]

4 0
3 years ago
M (b) Calculate the speed of an electron that is accelerated through the same electric potential difference.
MaRussiya [10]

The speed of a electron that is accelerated from rest through an electric potential difference of 120 V  is 6.49\times10^6m/s

<h3>How to calculate the speed of the electron?</h3>

We know, that the  energy of the system is always conserved.

Using the Law of Conservation of energy,

\triangle K+\triangle U=0

Here, K is the kinetic energy and U is the potential energy.

Now, substituting the formula of U and K, we get:

(\frac{1}{2} mv^2)+(-q\triangle V) =0------(1)

Here,

m is the mass of the electron

v is the speed of the electron

q is the charge on the electron

V is the potential difference

Let v_f and v_i represent the final and initial speed.

Here, v_i =0

Solving for v_f, we get:

v_f=\sqrt{\frac{-2q\triangle V}{m} }

=\sqrt{\frac{2\times1.602\times10^{-19}\times 120}{9.109\times10^{-31}} }

=6.49\times10^6m/s

To learn more about the conservation of energy, refer to:

brainly.com/question/2137260

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