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Lelechka [254]
3 years ago
11

A delivery man is told to go 60 miles per hour north to make his delivery on time. These instructions describe the delivery man'

s
-speed
-velocity
-acceleration
-magnitude
Physics
2 answers:
sergeinik [125]3 years ago
5 0
Velocity here because it's speed and direction
Nonamiya [84]3 years ago
4 0

Answer:

Speed

Explanation:

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A ball is thrown straight down with a speed of Va2.00 from a building 40.00 meters high. How much S time passes before the ball
-Dominant- [34]

Answer:

Option C is the correct answer.

Explanation:

Considering vertical motion of ball:-

Initial velocity, u =  2 m/s

Acceleration , a = 9.81 m/s²

Displacement, s = 40 m

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    40 = 2 x t + 0.5 x 9.81 x t²

   4.9t²  + 2t - 40 = 0

   t = 2.66 s   or t = -3.06 s

So, time is 2.66 s.

Option C is the correct answer.

6 0
3 years ago
In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

      $=\sqrt{2 \times 9.8 \times 0.35}$      (h = 35 cm = 0.35 m)

      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

      $=\frac{0.175}{0.525}+0$

     = 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

 

3 0
2 years ago
Using Newton's third law of motion, explain what happens when you let an untied balloon go.
mash [69]

Answer:

balloon pushes you back

Explanation:

3rd Law: Every action has an equal and opposite reaction

So, when you let go of the balloon it's pushed forward so the balloon pushes you back

7 0
3 years ago
Read 2 more answers
A truck on a straight road starts from rest, accelerating at 2.00m/s^2 until it reaches a speed of 20.0m/s. Then the truck trave
Dennis_Churaev [7]

Answer:

Explanation:

a )

Time to reach the speed of 20 m/s with an acceleration of 2 m/s² can be calculated as follows .

v = u + a t

20 = 0 + 2 t

t = 20 /2 = 10 s .

Total time = 10 s + 20 s + 5 s = 35 s .

b) Average velocity = Total distance travelled / total time

Distance travelled in first 10 s

S₁ = ut + 1/2 a t²

= 0 + .5 x 2 x 10²

= 100 m

Distance travelled in next 20 s

S₂= 20s x 20 m/s  = 400 m

Distance travelled in last 5 s .

deceleration in last 5 s

v = u + at

0 = 20 m/s + a x 5

a = - 4 m/s²

v² = u² - 2 a s

0 = (20 m/s)² - 2 x 4 m/s² x s

s = 50 m

S₃ = 50 m

Total distance = S₁ + S₂ + S₃

= 100 m + 400 m + 50 m

= 550 m .

Average velocity = 550 m / 35 s

= 15.71 m /s .

3 0
3 years ago
Satellites can focus on specific latitudes using:
Fittoniya [83]
B- east west orbits
4 0
2 years ago
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