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lord [1]
3 years ago
12

If a train travels at a speed of 40 km/h for 3 hours, how far will it travel

Physics
2 answers:
VikaD [51]3 years ago
8 0
120 km/3 hours. 40/1=?/3 1x3=3 hours so 40x3=120 km
Alik [6]3 years ago
4 0

Answer:

120 meters

Explanation:

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dedylja [7]

Answer:

A

Explanation:

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In general, an organism will be more likely to develop phobias of __________.
BARSIC [14]

Answer:

a) dangers faced during natural circumstances

7 0
2 years ago
please help me with my question I will like and mark as brainliest NO LINKS THEY DON'T WORK AND IF U DON'T KNOW THE ANSWER PLS D
bearhunter [10]

Answer:

a)

Weight in Air = 0.3N

Weight in Water = 0.25N

Weight in Liquid = 0.24N.

Upthrust /Buoyant Force = Weight in Air – Weight in Fluid(Water in this case)

= 0.3 – 0.25

= 0.5N.

b) R.D of Body = Density of Body/Density of Standard Fluid(Water).

There's a Derived Formula for RD.

I'm gonna Apply it here.

Ask me for the derivation in the Comment section if you need it.

RD = α/ρ = (Weight in Air) / (Upthrust Force)

Where

α = density of the Body(or reference substance)

ρ = density of standard fluid (water)

= 0.3/0.05 = 6.

c) RD of Liquid = (Density of Liquid) /(Density of standard Fluid(water)

Or we just go by that formula

RD of Liquid = Weight in Air/Upthrust(In Liquid)

We'll be using the Upthrust in that Liquid now.

= 0.3 – 0.24 = 0.06

RD = 0.3/0.06 = 5.

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3 years ago
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snow_tiger [21]
Soory i think multiple of velocity and angle
6 0
3 years ago
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
4 years ago
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