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Inessa [10]
4 years ago
14

Predict what would happen if earth’s plates stopped moving.

Physics
1 answer:
Hatshy [7]4 years ago
4 0
We would have earthquakes and tsunamis, i think, i don't really know?!?
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A car initially traveling at 60 km/h accelerates at a constant rate of 2.0 m/s2. how much time is required for the car to reach
bogdanovich [222]
<span>since the car starts at 60 km/hr, first subtract the 60 from 90 km/hr which is 30. that is the acceleration we need to solve. 2.0 m/s multiply by 60 seconds in a minute, and 60 minutes in an hour, which equals 7200 m/s in an hour or divide by 1000 (1000 meters in a km) and have 7.2m an hour. divide the 30 km/hr by 7.2 and you have 4.167 hours. You could also take the 30 km/hr needed to accelerate to match the 90 km/hr and find the number of meters needed to accelerate which is 30,000. divide by 2 meters per second equals 15,000. divide by 60 seconds in a minute, and divide by 60 minutes in an hour which still equals 4.167 hours.</span>
3 0
4 years ago
how much energy would microraptor gui have to expend to fly with a speed of 10 m/sm/s for 1.0 minute?
Ray Of Light [21]

Much energy as would Microraptor gui have to expend to fly with a speed of 10 m/s for 1.0 minutes is 486 J.

The first step is to find the energy that Microraptor must release to fly at 10 m/s for 1.0 minutes. The energy that Microraptor must expend to fly can be found using the relationship between Power and Energy.

P = E/t

Where:

P = power (W)

T = time (s)

Now, a minimum of 8.1 W is required to fly at 10 m/s. So, the energy expended in 1 minute (60 seconds) is

P = E/t

E = P x t

E = 8.1 x 60

E = 486 Joules

Thus, the energy that Microraptor must expend to fly at 10 m/s for 1.0 minutes is the 486 J.

Learn more about Microraptor gui here brainly.com/question/1200755

#SPJ4

3 0
1 year ago
What is the name of the force that acts between any two objects because of<br>their masses?​
svetoff [14.1K]

Answer:

Explanation:

The force of attraction between 2 masses.

4 0
4 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
What is the peak emf generated (in V) by rotating a 1000 turn, 42.0 cm diameter coil in the Earth's 5.00 ✕ 10−5 T magnetic field
ELEN [110]

Answer:

5.77 Volt

Explanation:

N = 1000, Diameter = 42 cm = 0.42 m,  t = 12 ms = 12 x 10^-3 s

Change in magnetic field, B = 5 x 10^-5 T

The peak value of emf is given by

e = N x dФ / dt

e = N x A x dB/dt

e = (1000 x 3.14 x 0.21 x 0.21 x 5 x 10^-5) / (1.2 x 10^-3)

e = 5.77 Volt

3 0
3 years ago
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