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Igoryamba
3 years ago
7

Simpson drives his car with an average velocity of 85 km/h eastward. How long will it take him to drive 560 km on a perfectly st

raight highway?
Physics
2 answers:
Nutka1998 [239]3 years ago
7 0

hahahahahahhha

wag kayo umasa sa brainly

mestny [16]3 years ago
6 0
560/85= 6.6 hr (2sf) = 6.6 * 3600 s = 2.4*10^4 s (2sf)
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An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t
Gelneren [198K]

Answer:

1650\:\mathrm{m}

Explanation:

We can use the following kinematics equations to solve this problem:

v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x.

Using the first one to solve for acceleration:

132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}.

Now we can use the second equation to solve for the distance travelled by the airplane:

132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}(three significant figures).

6 0
3 years ago
A camcorder has a power rating of 10 watts. If the output voltage from its battery is 3 volts, what current does it use?
Gemiola [76]
The power is calculated using the following rule:
Power = voltage * current
We are given that:
power = 10 watts
voltage = 3 volts

Substitute with the givens in the above equation to get the value of the current as follows:
P = V*I
10 = 3*I
current (I) = 10/3 amperes
7 0
3 years ago
A basketball player standing up with the hoop launches the ball straight up with an initial velocity of v_o = 3.75 m/s from 2.5
denis23 [38]

Answer:

a) The maximum height the ball will achieve above the launch point is 0.2 m.

b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.

Explanation:

a)

For the height reached, we use 3rd equation of motion:

2gh = Vf² - Vo²

Here,

Vo = 3.75 m/s

Vf =  0m/s, since ball stops at the highest point

g = -9.8 m/s² (negative sign for upward motion)

h = maximum height reached by ball

therefore, eqn becomes:

2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²

<u>h = 0.2 m</u>

b)

To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:

2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²

(Vo)² = 19.6 m²/s²

Vo = √19.6 m²/s²

<u>Vo = 4.43 m/s</u>

Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)

<u>Vo = 0.174 in/ms</u>

<u />

6 0
3 years ago
Not affected by acids a physical or chemical?
e-lub [12.9K]

Answer:

physical because chemical always reacts

3 0
3 years ago
Read 2 more answers
The work function of a metal surface is 4.80 × 10-19 J. The maximum speed of the electrons emitted from the surface is vA = 7.7
poizon [28]

Answer:

\lambda_A=2.65177\times 10^{-7}\ m

\lambda_B=3.32344\times 10^{-7}\ m

Explanation:

h = Planck's constant = 6.63\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

m = Mass of electron = 9.11\times 10^{-31}\ kg

W_0 = Work function = 4.8\times 10^{-19}\ J

v_A = Velocity of A particle = 7.7\times 10^5\ m/s

v_B = Velocity of B particle = 5.1\times 10^5\ m/s

The wavelength is given by

\lambda=\frac{hc}{\frac{1}{2}mv^2+W_0}

\lambda_A=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(7.7\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_A=2.65177\times 10^{-7}\ m

The wavelength \lambda_A=2.65177\times 10^{-7}\ m

\lambda_B=\frac{6.63\times 10^{-34}\times 3\times 10^8}{\frac{1}{2}9.11\times 10^{-31}(5.1\times 10^5)^2+4.8\times 10^{-19}}\\\Rightarrow \lambda_B=3.32344\times 10^{-7}\ m

The wavelength \lambda_B=3.32344\times 10^{-7}\ m

5 0
4 years ago
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