All categories

3 years ago

How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?

Physics

1 answer:

grandymaker [24]3 years ago

7 0

Answer:

-120000 W

Explanation:

Power = change in energy / time

P = ΔE / t

P = (½ mv₂² − ½ mv₁²) / t

P = m (v₂² − v₁²) / (2t)

Given m = 1.5 t = 1500 kg, v₂ = 10 m/s, v₁ = 30 m/s, and t = 5 s:

P = (1500 kg) ((10 m/s)² − (30 m/s)²) / (2 × 5 s)

P = -120000 W

You might be interested in

Urgently!

adelina 88 [10]

Answer:

0.003333 s to 0.000125s or from 3.33ms to 0.125ms wher m is for milli

1.1m to 0.04125 m

Explanation:

T= 1/f=

if f= 300Hz then T = 1/300 =0.003333 s

if f= 8000 then T= 1/8000 = 0.000125s

now v=f×wave length

or wavelength = speed/ frequency

when f = 300 Hz

wavelength = 330/300=1.1 m

wavelength = 330/8000 = 0.04125m

note : i have taken speed of sound as 330 m/s you can take any value given in between 330m/s to 340m/s

6 0

2 years ago

A 5.00 kg rock whose density is 4300 kg/m3 is suspended by a string such that half of the rock's volume is under water. You may

12345 [234]

Answer:

The tension on the string is $T = 43.302 \ N$

Explanation:

From the question we are told that

The mass of the rock is $m_r = 5.00 \ kg = 5000 \ g$

The density of the rock is $\rho = 4300 \ kg/m^3 = 4.3 g/dm^3$

Generally the volume of the rock is mathematically evaluated as

$V = \frac{m_r}{\rho}$

substituting values

$V = \frac{5000}{4.3}$

$V = 1162.7 \ dm^3$

The volume of the rock immersed in water is

$V_w = \frac{V}{2}$

substituting values

$V_w = \frac{1162.7 }{2}$

$V_w = 581.4 \ dm^3$

mass of water been displaced by the this volume is

$m_w = V_w$ According to Archimedes principle

=> $m_w = 581.4 \ g$

$m_w = 0.5814 \ kg$

The weight of the water displace is

$W _w = m_w * g$

$W _w = 0.5814 * 9.8$

$W _w = 5.698 \ N$

The actual weight of the rock is

$W_r = m_r * g$

$W_r = 5.0 * 9.8$

$W_r = 49.0 \ N$

The tension on the string is

$T = W_r - W_w$

substituting values

$T = 49.0 - 5.698$

$T = 43.302 \ N$

4 0

2 years ago

Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:

Natasha2012 [34]

The work done by $\vec F$ along the given path C from A to B is given by the line integral,

$\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r$

I assume the path itself is a line segment, which can be parameterized by

$\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath$

with 0 ≤ t ≤ 1. Then the work performed by F along C is

$\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}$

7 0

2 years ago

At 19 ºC a volume of air takes us 9.0 mL in an expandable container. What is its volume after it is heated up to 50. ºC?

Katena32 [7]

According to Charles law, we know, at constant pressure, volume is directly proportional to temperature.
So, V/T = constant

V₁/t₁ = V₂/t₂
V₁t₂ = V₂t₁

Here, we have: V₁ = 9 mL
V₂ = ?
T₂ = 50+272 = 323 K
T₁ = 19+273 = 292 K

Substitute their values into the expression:
9 × 323 = V₂ × 292
V₂ = 2907 / 292
V₂ = 9.95
After rounding-off to unit place value, it would be equal to 10 mL

So, In short Option C would be your correct answer.

Hope this helps!

5 0

2 years ago

Which statements about electric field lines are correct? Check all that apply.

slava [35]

Answer:

they cross over one another between charge.

7 0

3 years ago

How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?​

How much power should a braked 1.5t car have to be braked to reduce its speed from 30m / s to 10m / s in 5s?