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agasfer [191]
3 years ago
14

here ya go lol always with the warning "Don't use such phrases here, not cool! It hurts our feelings"

Physics
2 answers:
stiks02 [169]3 years ago
8 0

Answer:

thanks!

and i get that a lot- looks like we are bad guys

Explanation:

Vera_Pavlovna [14]3 years ago
5 0

Answer:

what

Explanation:

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A car that increase its speed from 20 km/h to 100 km/h undergoes -------acceleration,
stepladder [879]

negative acceleration- deceleration

8 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
2 years ago
A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?
Korvikt [17]

\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}

  • Speed of the mobile = 250 m/s
  • It starts decelerating at a rate of 3 m/s²
  • Time travelled = 45s

\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}

  • Velocity of mobile after 45 seconds

\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}

We can solve the above question using the three equations of motion which are:-

  • v = u + at
  • s = ut + 1/2 at²
  • v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}

We are provided with,

  • u = 250 m/s
  • a = -3 m/s²
  • t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>

<u>━━━━━━━━━━━━━━━━━━━━</u>

4 0
3 years ago
Find the velocity and distance travelled of a car that is accelerating at 3 m/s2 for 4 seconds.
marta [7]

Explanation:

Given:

v₀ = 0 m/s

a = 3 m/s²

t = 4 s

Find: Δx and v

Δx = v₀ t + ½ at²

Δx = (0 m/s) (4 s) + ½ (3 m/s²) (4 s)²

Δx = 24 m

v = at + v₀

v = (3 m/s²) (4 s) + 0 m/s

v = 12 m/s

6 0
2 years ago
A sample of chloroform is found to contain 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of
GREYUIT [131]

Given:

Sample 1:

Chloroform is CHCl_{3}

12 g Carbon

1.01 g Hydrogen

106.4 g Cl

Sample 2:

30.0 g of Carbon

Solution:

mass of chloroform from sample 1:

12 + 1.01 +106.4 =119.41 g

Now, for the total mass of chloroform in sample 2:

mass of chloroform \times\frac{given mass of carbon}{mass of carbon atom}

mass of chloroform = 119.41 \times\frac{30}{12} = 298.53 g

6 0
3 years ago
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