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1 year ago

I need help with nine and 10 I will really appreciate it

Physics

1 answer:

horsena [70]1 year ago

7 0

Answer:

9] V = D ÷ T

Take any distance value from the graph and its relevant time.

V = 4 ÷ 2

V = 2 m/s

[You will notice that any distance values with its time will give you 2 m/s as its speed. This means that speed is constant throughout.]

10] Take the distance value and its time for the highest peak of B.

V = 20 ÷ 2

V = 10 m/s

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A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to th

boyakko [2]

Answer:

8.76762 m

Explanation:

T = Time period = 5.94 seconds

g = Acceleration due to gravity = 9.81 m/s²

L = Length of pendulum = Height of tower

Time period is given by

$T=2\pi \sqrt{\dfrac{L}{g}}\\\Rightarrow T^2=4\pi^2\dfrac{L}{g}\\\Rightarrow L=\dfrac{T^2g}{4\pi^2}\\\Rightarrow L=\dfrac{5.94^2\times 9.81}{4\pi^2}\\\Rightarrow L=8.76762\ m$

The height of the tower is 8.76762 m

8 0

2 years ago

Review please help.

iragen [17]

Answer:

1 and 3

Explanation:

because they are going up from 0

7 0

2 years ago

a car travelling at 50km/h from rest covers a distance of 10km in 40minutes. Calculate the acceleration

Margarita [4]

Answer:

$9.67\cdot 10^{-3}m/s^2$

Explanation:

We can solve the problem by using the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem:

u = 0 (it starts from rest)

$v=50 km/h \cdot \frac{1000}{3600}=13.9 m/s$ is the final velocity

s = 10 km = 10 000 m is the displacement

Solving for a, we find:

$a=\frac{v^2-u^2}{2s}=\frac{13.9^2}{2(10000)}=9.67\cdot 10^{-3}m/s^2$

7 0

2 years ago

Match the following items.

Ray Of Light [21]

Answer:

Je ne Sachez que Qu’est-ce que le

8 0

2 years ago

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a

Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$ . Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$ . So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0

2 years ago