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3 years ago

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A 70kg man is moving with a speed of 3.2 m/s on a rough surface, his speed was decreasing uniformly until he reaches to a stop b

ecause of the friction of the rough surface. Calculate the friction force exerted on the man.

1 answer:

VMariaS [17]3 years ago

6 0

Answer:

224 N

Explanation:

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An ocean liner leaves New York City and travels 18.0o north of east for 155 km. How far east and how far north has it gone? In o

Monica [59]

I’m sorry if i took up a lot of space, hope this is a valid approximate answer

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1 year ago

A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 25.0 kg of water at 20.0°C. What is the final te

Ber [7]

Answer:

Te = 23.4 °C

Explanation:

Given:-

- The mass of iron horseshoe, m = 1.50 kg

- The initial temperature of horseshoe, Ti_h = 550°C

- The specific heat capacity of iron, ci = 448 J/kgC

- The mass of water, M = 25 kg

- The initial temperature of water, Ti_w = 20°C

- The specific heat capacity of water, cw = 4186 J/kgC

Find:-

What is the final temperature of the water–horseshoe system?

Solution:-

- The interaction of horseshoe and water at their respective initial temperatures will obey the Zeroth and First Law of thermodynamics. The horseshoe at higher temperature comes in thermal equilibrium with the water at lower temperature. We denote the equilibrium temperature as (Te) and apply the First Law of thermodynamics on the system:

m*ci*( Ti_h - Te) = M*cw*( Te - Ti_w )

- Solve for (Te):

m*ci*( Ti_h ) + M*cw*( Ti_w ) = Te* (m*ci + M*cw )

Te = [ m*ci*( Ti_h ) + M*cw*( Ti_w ) ] / [ m*ci + M*cw ]

- Plug in the values and evaluate (Te):

Te = [1.5*448*550 + 25*4186*20 ] / [ 1.5*448 + 25*4186 ]

Te = 2462600 / 105322

Te = 23.4 °C

7 0

3 years ago

Read 2 more answers

Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T

ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

$\vec{A}=x_A\hat{i}$

Vector B , which has a magnitude of 5.5 m

$\vec{B}=x_B\hat{i}+y_B\hat{j}$

$\sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25$

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A $\vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}$

Comparing we will get

$x_A=-x_B\\\\y_B=6x_A$

Substituting in $x_{B}^{2}+y_{B}^{2}=30.25$

$\left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904$

So we have

$\vec{A}=0.904\hat{i}$

Magnitude of vector A = 0.904

8 0

3 years ago

what is the answer to what does a 150 kg roller coaster traveling at 10m/s has how much kinetic energy

Svetllana [295]

The answer to the question is 15 kinetic energy

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3 years ago

The magnetic field over a certain range is given by B~ = Bx ˆı + By ˆ, where Bx = 2 T and By = 4 T. An electron moves into the

krek1111 [17]

Answer:

Explanation:

The force exerted in a magnetic field is given as

F = q (v × B)

Where

F is the force entered

q is the charge

v is the velocity

B is the magnetic field

Given that,

The magnetic field is

B = 2•i + 4•j. T

The velocity of the electron is

v = 2•i + 6•j + 8•k. m/s

Also, the charge of an electron is

q = -1.602 × 10^-19 C.

Then note that,

V×B is the cross product of the speed and the magnetic field

Then,

F = q (V×B)

F = -1.602 × 10^-19( 2•i + 4•j +8•k × 2•i + 4•j)

Note

i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

F = -1.602 × 10^-19[(2•i + 4•j +8•k) × (2•i + 4•j)]

F = -1.602 × 10^-19 [2×2•(i×i) + 2×4•(i×j) + 4×2•(j×i) + 4×4•(j×j) + 8×2•(k×i) + 8×4•(k×j)]

F = -1.602 × 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]

F = -1.602 × 10^-19(0 + 8•k - 8•k + 0 + 16•j - 32•i)

F = -1.602 × 10^-19(16•j - 32•i)

F = -1.602 × 10^-19 × ( -32•i + 16•j)

F = 5.126 × 10^-18 •i - 2.563 × 10^-18 •j

Then, the x component of the force is

Fx = 5.126 × 10^-18 N

Also, the y component of the force is

Fy = -2.563 × 10^-18 N

8 0

3 years ago