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creativ13 [48]
3 years ago
10

A car is driving across a street. It's change in displacement is 50 m. If it started with a velocity of 5 m/s and ended with a v

elocity of 7 m/s, what must the acceleration have been?
Physics
1 answer:
MrMuchimi3 years ago
8 0

Answer:

Ew an sao fan. anyway, it's 0.24m/s^2

Explanation:

Use this equation for 1 dimensional motion with constant acceleration:

Xf = Xi + Vi*T + 0.5A*T^2

Xf is 50 m when Xi is 0m

Vi is 5m/s

A is (7m/s-5m/s)/T = 2m/s/T

So,

50m = 0m + 5m/s *T + 0.5*2m/s/T * T^2

50m = 5m/s * T + 1m/s * T

50m = 6m/s * T

T = 8.33s

A is 2m/s/T

A is 2m/s/8.33s

A is 0.24m/s^2

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The density of blood is 2kg/m.convert to g/cm
sertanlavr [38]

Explanation:

if we convert it into g/cm it'll be

2000 grams

6 0
2 years ago
Calculate the magnitude of electric field strength at a point 3cm from an infinite line of charge of linear density 18 μC/cm sit
valina [46]
Answer:

The magnitude of the electric field strength = 7.2 x 10⁸ N/C

Explanation:

The linear density:

\begin{gathered} \lambda=18\mu C\text{ /cm} \\  \\ \lambda=\frac{18*10^{-6}C}{0.01m} \\  \\ \lambda=0.0018\text{ C/m} \end{gathered}

Point r = 3 cm = 3/100 m

r = 0.03 m

The electric field strength is calculated below

\begin{gathered} E=\frac{\lambda}{2\pi\epsilon_o\epsilon_rr} \\  \\ E=\frac{0.0018}{2\times3.14\times8.85\times10^{-12}\times1.5\times0.03} \\  \\ E=719709237.468\text{ N/C} \\  \\ E=7.2\times10^8\text{ N/C} \\  \end{gathered}

The magnitude of the electric field strength = 7.2 x 10⁸ N/C

8 0
1 year ago
I need help on this!
sergij07 [2.7K]
Try looking it up. that might help
8 0
3 years ago
A rock is dropped from the top of a tower. When it is 40 meters above the ground velocity of 17 m/s. When its velocity is 24 m/s
eimsori [14]

Answer:

Option B is the correct answer.

Explanation:

Let us consider 40 meter above ground as origin.

Initial velocity = 17 m/s

Final velocity = 24 m/s

Acceleration = 9.81 m/s

We have equation of motion v² = u² + 2as

Substituting

         24² = 17² + 2 x 9.81 x s

           s = 14.63 m

Distance traveled by rock = 14.63 m down.

Height of rock from ground = 40 - 14.63 = 25.37 m = 25.4 m

Option B is the correct answer.

4 0
3 years ago
A star with the same mass and diameter as the sun rotates about a central axis with a period of about 24.0 days. Suppose that th
Mila [183]

Answer:

a)  w = 2.52 10⁷ rad / s, b)  K / K₀ = 1.19 10⁴

Explanation:

a) We can solve this exercise using the conservation of angular momentum.

Initial instant. Before collapse

         L₀ = I₀ w₀

Final moment. After the collapse

         L_f = I w

angular momentum is conserved

        L₀ = L_f

         I₀ w₀ = I w                 (1)

         

The moment of inertia of a sphere is

        I = 2/5 m r²

we take from the table the mass and diameter of the star

        m = 1,991 10³⁰ kg

        r₀ = 6.96 10⁸ m

        r = 6.37 10⁶ m

to find the angular velocity let's use

       w = L / T

where the length of a circle is

      L = 2π r

      T = 24 days (24 h / 1 day) (3600 s / 1h) = 2.0710⁶ s

we substitute

      w = 2π r / T

      wo = 2π 6.96 10⁸ / 2.07 10⁶

      wo = 2.1126 10³ rad / s

we substitute in equation 1

      w = \frac{I_o}{I}

      w = 2/5 mr₀² / 2/5 m r² w₀

      w = (\frac{r_o}{r}) ² wo

      w = (6.96 10⁸ / 6.37 10⁶) ² 2.1126 10³

      w = 2.52 10⁷ rad / s

b) the kinetic energy ratio

      K = ½ m w²

       K₀ = ½ m w₀²

       K = ½ m w²

       K / K₀ = (w / wo) ²

       K / K₀ = 2.52 10⁷ / 2.1126 10³

       K / K₀ = 1.19 10⁴

7 0
3 years ago
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