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3 years ago

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A driver in a car traveling at a speed of 26.8 m/s sees a deer 100 m away on the road. Calculate the minimum constant accelerati

on that is necessary of the car to stop without hitting the deer (assuming the deer does not move).
-4.20 m/s2

-1.75 m/s2

-2.89 m/s2

-3.59 m/s2

1 answer:

ZanzabumX [31]3 years ago

6 0

-3.5912m/s^2Use vf^2 = vi^2 + 2ax in your calculations, knowing vf, vi, and x.

You might be interested in

Please help, and show steps. Thank you very much!

Vikentia [17]

V = 8 * 10^2 km/h = 800km/h
S= 1,8* 10^3 km = 1800km
t = ?
v = S/t
t = S/v
t = 1800km/ 800km/h
t ≈ 2,25h (135min)

6 0

3 years ago

The slope of a velocity versus time graph gives

Marina86 [1]

Explanation:

Average of acceleration

4 0

3 years ago

A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west

just olya [345]

Answer:

5 m/s

Explanation:

Given that,

A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west.

It is assumed to find the resultant velocity of the vehicle. Let east side is positive and west is negative. So,

$v=v_1+v_2\\\\=20+(-15)\\\\=5\ m/s$

Hence, the resultant velocity of the vehicle is equal to 5 m/s.

6 0

3 years ago

A 75.00 gram sample of an unknown metal initially at 99.0 degrees Celcius is added to 50.00 grams of water initially at 10.79 de

jeyben [28]

Answer:

$c_{e1}$ = 0.331 J / g ° C

Explanation:

We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.

Heat ceded Qh = m1 ce1 ( $T_{h}$ - $T_{f}$ )

Heat absorbed Qc = m2 ce2 ( $T_{f}$ - T₀)

Body 1 is metal and body 2 is water . Where m are the masses of the two bodies, ce their specific heat and T the temperatures

Qh = Qc

m₁ $c_{e1}$ ( $T_{h}$ - $T_{f}$ ) = m₂ $c_{e2}$ ( $T_{f}$ - T₀)

we clear the specific heat of the metal

$c_{e1}$ = m₂ $c_{e2}$ ( $T_{f}$ - T₀) / (m₁ ( $T_{h}$ - $T_{f}$ ))

$c_{e1}$ = 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))

$c_{e1}$ = 209.2 (9.36) / (75 78.85)

$c_{e1}$ = 1958.11 / 5913.75

$c_{e1}$ = 0.331 J / g ° C

5 0

3 years ago

Assuming that the tungsten filament of a lightbulb is a blackbody, determine its peak wavelength if its temperature is 3 200 K.

rosijanka [135]

Answer:

the peak wavelength when the temperature is 3200 K = $9.05625*10^{-7} \ m$

Explanation:

Given that:

the temperature = 3200 K

By applying Wien's displacement law ,we have

$\lambda _m$ T = 0.2898×10⁻² m.K

The peak wavelength of the emitted radiation at this temperature is given by

$\lambda _m$ = $\frac{0.2898*10^{-2} m.K}{3200 K}$

$\lambda _m$ = $9.05625*10^{-7} \ m$

Hence, the peak wavelength when the temperature is 3200 K = $9.05625*10^{-7} \ m$

6 0

3 years ago