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spayn [35]
3 years ago
13

A bicyclist moves at a constant speed of 6 m/s. How long it will take for the point bicyclist to move 36 m?​

Physics
2 answers:
GenaCL600 [577]3 years ago
6 0

Answer:

Somuchcringe answered correctly you mark him brainliest

Explanation:

Rudik [331]3 years ago
3 0

Answer:

6 seconds

Explanation:

make me brainliest

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If you weigh 637 N on Earth's surface, how much would you weigh on the planet Mars?
vivado [14]

Answer:

To get your answer multiply 637N by 10N/kg to get 6370kg

4 0
2 years ago
A runner starts from rest and achieves a maximum speed of 8.85 m/s.
ycow [4]

Answer:

The appropriate answer is "0.9152 seconds".

Explanation:

The given values are:

Maximum speed,

v = 8.85 m/s

Acceleration,

u = 9.67 m/s²

Now,

⇒ v = u+at

By putting the values, we get

⇒ 8.85=0+9.67t

⇒      t=\frac{8.85}{9.67}

⇒         =0.9152 \ second    

5 0
2 years ago
Imagine a 15kg block moving with a speed of 20m/s. Calculate the kinetic energy of this block. (Show the equation, show your wor
dmitriy555 [2]

Answer:

The answer to your question is:        Ke = 3000 Joules

Explanation:

Data

mass = 15 kg

speed = 20 m/s

Kinetic energy = ?

Equation

               Ke = \frac{1}{2}mv^{2}

               Ke = \frac{1}{2}(15)(20)^{2}

                Ke = 3000 Joules

5 0
3 years ago
Read 2 more answers
Who thinks this a ufo? The following images are from a rare footage.
Galina-37 [17]

Answer:

That might be a sled ngl but it looks like a ufo but it's definitely not

5 0
2 years ago
Read 2 more answers
Please help thank you
Margarita [4]

Answer:

\theta \approx 59.036^{\circ}, T_{2} \approx 23.324\,N

Explanation:

First we build the Free Body Diagram (please see first image for further details) associated with the mass, we notice that system consist of a three forces that form a right triangle (please see second image for further details): (i) The weight of the mass, (ii) two tensions.

The requested tension and angle can be found by the following trigonometrical and geometrical expressions:

\theta = \tan^{-1} \frac{W}{T_{2}} (1)

T_{1} = \sqrt{W^{2}+T_{2}^{2}} (2)

Where:

W - Weight of the mass, measured in newtons.

T_{1}, T_{2} - Tensions from the mass, measured in newtons.

If we know that W = 20\,N and T_{2} = 12\,N, then the requested values are, respectively:

\theta = \tan^{-1} \frac{20\,N}{12\,N}

\theta \approx 59.036^{\circ}

T_{2} = \sqrt{(20\,N)^{2}+(12\,N)^{2}}

T_{2} \approx 23.324\,N

7 0
2 years ago
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