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avanturin [10]
3 years ago
7

complete the analogy: 1. speed: velocity;______:______ a. acceleration:velocity b. distance:displacement c. speed:acceleration d

. time:scalar 2. positive acceleration:_____: zero acceleration:constant velocity a. constant speed b. decreasing velocity c. increasing velocity d. zero velocity
Physics
2 answers:
Paraphin [41]3 years ago
7 0

Velocity is the vector,
     made up of the scalar speed + direction of speed

b).  Displacement is the vector,
           made up of the scalar distance + direction of distance
maks197457 [2]3 years ago
6 0
B.
distance is used to calculate speed

displacement is used to calculate velocity

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A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting
ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

T=\dfrac{38}{8}\\\\T=4.75\ s

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

k = spring stiffness constant

So,

k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m

When the cord is in air,

mg=kx

x = the extension in the cord

x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0
3 years ago
Do you like genetic engineering explain even if there is no answer box still answer in the commets
liubo4ka [24]

Answer:

yes it is so awesome

Explanation:

4 0
3 years ago
In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
muminat

Answer:

2.1\cdot 10^{21} electrons

Explanation:

The magnitude of the electric field outside an electrically charged sphere is given by the equation

E=\frac{kQ}{r^2}

where

k is the Coulomb's constant

Q is the charge stored on the sphere

r is the distance (from the centre of the sphere) at which the field is calculated

In this problem, the cloud is assumed to be a  charged sphere, so we have:

E_b=3.00\cdot 10^6 N/C is the maximum electric field strength tolerated by the air before breakdown occurs

r=1.00 km = 1000 m is the radius of the sphere

Re-arranging the equation for Q, we find the maximum charge that can be stored on the cloud:

Q=\frac{Er^2}{k}=\frac{(3.00\cdot 10^6)(1000)^2}{9\cdot 10^9}=333.3 C

Assuming that the cloud is negatively charged, then

Q=-333.3 C

And since the charge of one electron is

e=-1.6\cdot 10^{-19}C

The number of excess electrons on the cloud is

N=\frac{Q}{e}=\frac{-333.3}{-1.6\cdot 10^{-19}}=2.1\cdot 10^{21}

5 0
2 years ago
An auditorium measures 40.0 m 3 20.0 m 3 12.0 m. The density of air is 1.20 kg/m3. What are (a) the volume of the room in cubic
Burka [1]

We use the formula,

m = V\rho

Here, m is the mass, V is the volume and  \rho density

Also

V = l w h

Here l is length, w is width and h is height.

(a) The volume of the room,

V=40.0 \ m \times 3 20.0 \ m \times 3 12.0\ m = 3.99 \times 10^{6} m^3

The volume of the room in cubic feet,

V = 3.99 \times 10^{6} m^3 \times(\frac{3.281 \ ft}{m} )^3 = 4.3 \times 10^7 \ ft^3

(b) Now the mass of the air in room,

m= (3.99 \times 10^{6} \ m^3) (1.20 \ kg/m^3) = 4.8 \times 10^6 kg.

Therefore, the weight of the air in room,

W = mg= 4.8 \times 10^6 kg \times 9.8 m/s^2 = 46.9 \times 10^6 \ N \\\\ W = 4.69 \times 10^7 \ N.

The weight of air in the room in pounds,

W = 4.69 \times 10^7 \ N ( \frac{1 \ lb}{4.448 \ N} ) = 1.1 \times 10^7 \ lb


3 0
3 years ago
A 22-turn circular coil of radius 3.00 cm and resistance 1.00 Ω is placed in a magnetic field directed perpendicular to the plan
zloy xaker [14]

Answer:

23.5 mV

Explanation:

number of turn coil  'N' =22

radius 'r' =3.00 cm=> 0.03m

resistance = 1.00 Ω

B= 0.0100t + 0.0400t²

Time 't'= 4.60s

Note that Area'A' = πr²

The magnitude of induced EMF is given by,

lƩl =ΔφB/Δt = N (dB/dt)A

    =N[d/dt (0.0100t + 0.0400 t²)A        

    =22(0.0100 + 0.0800(4.60))[π(0.03)²]

     =0.0235

     =23.5 mV

Thus, the induced emf in the coil at t = 4.60 s is 23.5 mV

8 0
3 years ago
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