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8090 [49]
3 years ago
8

A floating ice block is pushed through a displacement d = (13.5 m)i + (-14.3 m)j along a straight embankment by rushing water, w

hich exerts a force f = (230 n)i + (-165 n)j on the block. how much work does the force do on the block during the displacement?
Physics
1 answer:
fredd [130]3 years ago
7 0

To solve this problem, we are going to use the formula for work which is Fd where x and y are measured separately.

 

X direction: W = 13.5 x 230 = 3105 Joules

Y direction: W = -14.3 x -165 = 2360 Joules

So the total work is getting the sum of the two: 3105 + 2360 = 5465 Joules

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4 0
2 years ago
List the submultiples and multiple units of length, mass, and time with respect to real-life situations. How are these units are
dangina [55]

Answer:

Explanation:

In physics, there are two types of physical quantities namely the fundamental and the derived quantities. Fundamental quantities are independent quantities on which derived quantities depends on. Length, mass and time are examples of fundamental quantities.

The SI unit of length is meters. A meter is a multiple unit. Its submultiple units are centimetres (10⁻²metres), kilometres (10³metres), decimetres (10⁻¹metres) etc

The SI unit of mass is kilogram (kg). The only sub multiple unit used in real-life situation is grams.

1 kg = 100 grams

The SI unit of time is seconds. The multiple units are the minutes, hours, weeks, days and years.

1 minute = 60 seconds

1 hour = 3600 seconds

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3 years ago
A 25.0 kg bag of peat moss sits in the back of a flatbed truck, driving up a hill. The bag experiences a 225N normal force. The
malfutka [58]

Answer:

a

   \theta  =  23.32^o

b

  \mu_s =  0.27

c

s =  0.948 \  m

Explanation:

From the question we are told that

The mass of the bag is m_b  =  25.0 \  kg

The normal force experienced is F_n  =  225 \ N

The maximum acceleration of the bag is a =  2.40 \  m/s^2

Generally this normal force experience by the bag is mathematically represented as

F_n  =  mg cos \theta

=> 225  =  (25 * 9.8) cos \theta

=> 0.9183  =   cos \theta

=> \theta  = cos^{-1}[0.9183]

=> \theta  =  23.32^o

Generally for the bag not to slip , it means that the frictional force is equal to the sliding force

F_f =  F_s

Hence F_f is mathematically represented as

F_f   =  \mu_s  *  F_n

While F_s is mathematically represented as

F_s   =  m * a

So

\mu_s  *  F_n = m * a

=> \mu_s  *  225 = 25 * 2.40

=> \mu_s =  0.27

Generally from the workdone equation we have that

KE_f - KE_i =  W_f

Here W_f is the work done by friction which is mathematically represented as

W_f  =  m * g * \mu_k * s

Here s is the distance covered by the bag

KE_f is zero given that velocity at rest is zero

and

KE_i = \frac{1}{2}  *  m* v_i^2

so

   \frac{1}{2}  *  m* v_i^2 = m * g * \mu_k * s

=>  \frac{1}{2}  *  v_i^2 =   g * \mu_k * s

substituting  2.55 m/s for v_i and 0.350 for  \mu_k  we have that

     \frac{1}{2}  *  2.55^2 =   9.8 * 0.350 * s

=> s =  0.948 \  m

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Who what where when why how ?
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When using the magnification equation, a value greater than 1 as the solution for M indicates that the image ? is larger than th
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M= Height of image/height of the object

If, M>1, then height of image>height of object.

And therefore, image is larger than the object.
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