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kirill115 [55]
3 years ago
10

-5 1/3 - 2 1/3 in math

Physics
2 answers:
Angelina_Jolie [31]3 years ago
8 0
-7 2/3 or -23 / 3

Hope this helped :)
velikii [3]3 years ago
7 0

Answer:

3

Explanation:

5 1/3 - 2 1/3  = 3

Broken down: 1/3 - 1/3 = 0      5 - 2 = 2

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A motorcycle has a mass of 2.50×10^2. A constant force is exerted on it for 60.0s. The motorcycles initial velocity is 6.00 m/s
Margarita [4]

The change in momentum is 5500 kg m/s

Explanation:

The change in momentum of an object is given by

\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have:

m=2.50\cdot 10^2 kg (mass of the motorcycle)

v=28.0 m/s (final velocity)

u=6.0 m/s (initial velocity)

Therefore, the change in momentum is

\Delta p=(2.50\cdot 10^2)(28.0-6.0)=5500 kg m/s

Learn more about change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

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4 years ago
Evaluate (x +y)0 for x= -3 and y=5.<br> 0 1<br> 2<br> 01
frosja888 [35]

Answer:2.01201

Explanation:

8 0
3 years ago
A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet
astraxan [27]

Answer:

Acceleration = 2.35 m/s^{2}

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

Friction along incline = umg cos30° = mg 0.3\times \frac{\sqrt{3}}{2}

Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

Length of the inclined edge = 16 m

v^{2}=u^{2}+2as

v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

5 0
4 years ago
The force applied when using a simple machine
jenyasd209 [6]

Answer:

effort

Explanation:

effort is the answer

7 0
4 years ago
Read 2 more answers
8
aev [14]
OD because Boyle’s law specifically states
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3 years ago
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