An eagle flies from its perch in a tree to the ground to capture and eat its prey. describe its energy transfromation

20 POINTS 20 POINTS On a bright sunny day you decide to take a walk. You begin at your home and walk 1000 meters to an ice cream

Kay [80]

Answer:

A) Total Distance = 2000 m and Total displacement = 0 m

B) Average Speed = 44.44 m/min and Average Velocity = 0 m/min

C) Average Speed = 0.7407 m/s and Average velocity = 0 m/s

Explanation:

A) Distance to reach ice cream shop from home = 1000 meters

Therefore distance to get back home would also be 1000 meters.

Total distance traveled = 1000 + 1000 = 2000 metres

Since journey started at home and ended at home, then total displacement = 0 metres.

B) Average speed = Total distance/total time.

Total time = 10 + 15 + 20 = 45 minutes

Since total distance = 2000 m

Then;

Average speed = 2000/45

Average speed = 44.44 m/min

Average velocity = Total displacement/total time

Average velocity = 0/45 = 0 m/min

C) We now want answers in B to be in m/s.

Total time = 45 minutes.

From conversion, 60 seconds make 1 minute. Thus, 45 minutes = 45 × 60 = 2700 seconds

Thus;

Average Speed = 2000/2700

Average Speed = 0.7407 m/s

Average displacement = 0/2700 = 0 m/s

4 0

3 years ago

Two students walk in the same direction along a straight path at a constant speed. One walks at a speed of 0.90 m/s and the othe

mr Goodwill [35]

Answer:

A. 456 seconds

Explanation:

We are given that two students walk in the same direction along a straight path at a constant speed.

One student walks with a speed=0.90 m/s

second student walks with speed=1.9 m/s

Total distance covered by each students=780 meter

We have to find who is faster and how much time extra taken by slower student than the faster student.

Time taken by one student who travel with speed 0.90 m/s= $\frac{780}{0.90}$

Time= $\frac{distance}{speed}$

Time taken by one student who travel with speed 0.90 m/s

= $\frac{780}{0.90}$

Time taken by one student who travel with speed 0.90 m/s

=866.6 seconds

Time taken by second student who travel with speed 1.9 m/s= $\frac{780}{1.9}$

=410.5 seconds

The second student who travels with speed 1.9 m/s is faster than the student travels with speed 0.90 m/s .

Extra time taken by the student travels with speed 0.90 m/s=866.6-410.5=456.1 seconds

Extra time taken by the student travels with speed 0.90 m/s=456 seconds

Hence, option A is true.

7 0

2 years ago

Read 2 more answers

If the battery of your phone can provide 2 mA of current to your phone and holds a charge of 130 C, how long will it take a full

Naya [18.7K]

Answer: 65000 seconds

Explanation:

Given that,

Current (I) = 2 mA

(Since 1 mA = 1 x 10^-3A

2 mA = 2 x 10^-3A)

Charge (Q) = 130 C

Time taken for a fully charged phone to die (T) = ?

Recall that the charge is the product of current and time taken.

i.e Q = I x T

130C = 2 x 10^-3A x T

T = 130C / (2 x 10^-3A)

T = 65000 seconds (time will be in seconds because seconds is the unit of time)

Thus, it will take a fully charged phone 65000 seconds to die

5 0

3 years ago

A car of mass 1470 kg is on an icy driveway

Schach [20]

Answer:

a = 7.5 m / s²

Explanation:

For this exercise let's use Newton's second law, let's create a coordinate system with the x axis parallel to the plane and the y axis perpendicular to the plane

Y axis

N - W cos θ = 0

N = mg cos θ

X axis

W sin θ = m a

mg sin θ = m a

a = g sin θ

let's calculate

a = 9.8 cos 40

a = 7.5 m / s²

8 0

2 years ago

A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)

Alekssandra [29.7K]

Answer:

The electric field strength is $6.6\times10^{5}\ V/m$

Explanation:

Given that,

Magnetic field = 0.150 T

Speed $v= 4.40\times10^{6}\ m/s$

We need to calculate the electric field strength

Using formula of velocity

$v=\dfrac{E}{B}$

$E=v\times B$

Where, v = speed

B = magnetic field

Put the value into the formula

$E=4.40\times10^{6}\times0.150$

$E=660000\ V/m$

$E=6.6\times10^{5}\ V/m$

Hence, The electric field strength is $6.6\times10^{5}\ V/m$

4 0

2 years ago