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kolbaska11 [484]
3 years ago
10

An object’s position vs. time graph is a straight line with positive slope, draw the velocity vs. time graph that represents thi

s motion
Physics
1 answer:
maksim [4K]3 years ago
8 0

Answer:

Explanation:

its going to be a straight line with increasing acceleration  know that slop of velocity -tine graph is acceleration

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PLZ HELP WILL GIVE BRAINLIEST
xxTIMURxx [149]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of the diver = 77kg

Height  = 8.18m

Unknown:

Final velocity  = ?

Solution:

To solve this problem, we use one of the motion equations.

            v²  = u² + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

             v² = 0² + (2 x 9.8 x 8.18)

             v² = 160.3

             v = 12.7m/s

7 0
3 years ago
You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l
sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0
1 year ago
The resistance created by waves on a 120-m-long ship is tested in a channel using a model that is 4 m long Y Part A If the ship
DanielleElmas [232]

Answer:

V_m = 12.78 km/hr

Explanation:

given,

length of the ship = 120 m

length of model of the ship = 4 m

Speed at which the ship travels = 70 km/h

speed of model = ?

by using froude's law

  F_r = \dfrac{V}{\sqrt{L g}}

for dynamic similarities

  (\dfrac{V}{\sqrt{L g}})_P = (\dfrac{V}{\sqrt{L g}})_{model}

  (\dfrac{V_p}{\sqrt{L_p}}) = (\dfrac{V_m}{\sqrt{L_m}})

  (\dfrac{70}{\sqrt{120}}) = (\dfrac{V_m}{\sqrt{4}})

          V_m = 12.78 km/hr

hence, the velocity of model will be 12.78 km/h

6 0
3 years ago
The charges that are free to move in a metallic conducting wire and that are responsible for the flow of electric current are- a
SVEN [57.7K]

Answer:

D) The negatively charged electrons

Electricity passes through metallic conductors as a flow of negatively charged electrons. The electrons are free to move from one atom to another. We call them a sea of delocalised electrons. Current was originally defined as the flow of charges from positive to negative. Please give me the brainliest answer?

:) Hoped this helped!!! Have a good day!!! <3

3 0
2 years ago
A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?
katrin [286]

Answer:

<em>k = 25.18 N/m</em>

Explanation:

<u>Simple Harmonic Oscillator</u>

It consists of a weight attached to one end of a spring being allowed to move forth and back.

If m is the mass of the weight and k is the constant of the spring, the period of the oscillation is given by:

\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}

If the period is known, we can find the value of the constant by solving for k:

\displaystyle k=m\left(\frac{2\pi}{T}\right)^2

Substituting the given values m=5 Kg and T=2.8 seconds:

\displaystyle k=5 \left(\frac{2\pi}{2.8}\right)^2

k = 25.18 N/m

5 0
3 years ago
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