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Anastaziya [24]
3 years ago
6

Suppose you take a trip that covers 180 km and takes 3 hours to make. Your average speed is A. 30 km/h. B. 60 km/h. C. 180 km/h.

D. 360 km/h. E. 540 km/h.
Physics
1 answer:
Ivenika [448]3 years ago
5 0
B. 60 kilometers per hour
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In a typical lightning strike, 2.0 c flows from cloud to ground in 0.24 ms. what is the current during the strike?
lyudmila [28]
Current = charge/time = (2 c)/(0.00024 sec)= 8,333 Amps !
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Which term refers to a substance that can only be broken down chemically? A. metals B. molecules C. matter D. mixtures
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C Is what I believe the answer is
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The pressure in a compressed air storage tank is 1200 kPa. What is the tank’s pressure in (a) kN and m units; (b) kg, m, and s u
elixir [45]

Explanation:

Given data

Pressure P=1200 kPa

To find

Pressure in

(a) kN/m²

(b) kg/m.s²

(c) kg/km.s²

Solution

For Part (a)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )\\P=1200kN/m^{2}

For Part (b)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )\\P=1,200,000kg/m.s^{2}

For Part (c)

P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )(\frac{100m}{1km} )\\P=1,200,000,000kg/km.s^{2}

8 0
3 years ago
The things that make up a mixture ________ their individual properties.
PIT_PIT [208]

Answer: Share

Explanation: <em>A mixture in which its constituents are distributed uniformly is called homogeneous mixture, such as salt in water. A mixture in which its constituents are not distributed uniformly is called heterogeneous mixture, such as sand in water.</em>

5 0
3 years ago
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Bob is standing at rest. Wendy runs past him at a constant speed of 3.00 m/s. At the instant that Wendy passes him, Bob starts t
77julia77 [94]

Answer:

90 meters

Explanation:

To find the distance in which Bob catches up with Wendy, you first write down the motion equations of both Bob and Wendy.

Wendy has a constant speed, then you have:

x_1=v_1t           (1)

Bob has an accelerated motion, then, his equation of motion is:

x_2=v_2t+\frac{1}{2}at^2          (2)

v1: constant speed of Wendy = 3.00 m/s

v2: initial speed of Bob = 0 m/s

a: acceleration of Bob = 0.200m/s^2

When Bob catches up with Wendy x1 = x2, then you equal both equations and solve for time t:

x_1=x_2\\\\v_1t=\frac{1}{2}at^2\\\\t=\frac{2v_1}{a}

you replace the values of a and v1:

t=\frac{2(3.00m/s)}{0.200m/s^2}=30 s

Finally, you replace this value of time either the equation for x1 or the equation for x2, and you calculate the distance:

x_2=x_1=(3.00m/s)(30s)=90m

hence, Bob catches up with Wendy for a distance of 90m

3 0
3 years ago
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