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3 years ago

What is the mass of an object with 15,000 J of kinetic energy moving at 20 m/s?

Physics

1 answer:

insens350 [35]3 years ago

6 0

Answer:

kE=1/2mv^2

Explanation:

here

KE= 15000j

velocity =20m/s

mass=?

now

mass=

15000×2= (m)20^2

30000=m ×400

30000÷400=m

Mass (m)= 75

therefore mass is 75kilogram

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Describe the evolution of a pulsar over time, in particular how the rotation and pulse signal changes over time.

madam [21]

Answer:

As beams of particles and their associated energy are given off, the pulsar will lose energy slowly, which will decrease the rate of its rotation. The frequency of pulses would therefore decrease, so that fewer pulses are observed in a given time span. The strength of the pulse signal will also decrease so the pulses will become fainter. Eventually, the pulsar should rotate so slowly and have such a low emission of radiation that it would no longer be observable.

3 0

4 years ago

7. These temperatures were recorded in Pasadena for a week in April. 87 85 80 78 83 86 90 Find each of these. (a) Mean (e) Range

slega [8]

Answer:

a) Mean = 84.14

b) Median = 85

c) Mode = no mode (since there is no variable that appears more than once in this dataset)

d) Midrange = 84

e) Range = 12

f) Variance = 14.69

g) Standard deviation = 3.83

Explanation:

The raw data to be processed is

87 85 80 78 83 86 90

a) Mean = (Σx)/N

The mean is the sum of variables divided by the number of variables

x = each variable

N = number of variables = 7

Mean = (87+85+80+78+83+86+90)/7

Mean = 84.14

b) Median is the number in the middle of the dataset when the variables are arranged in ascending or descending order.

Arranging the data in ascending order

78, 80, 83, 85, 86, 87, 90

The number in the middle is the 4th number = 85

Median = 85

c) Mode is the variable that occurs the most in a distribution.

For this question, all of the variables occur only once, with no variable occurring more than once. Hence, there is no mode for this dataset.

d) Midrange is the arithmetic mean of the highest and lowest number in the dataset.

Mathematically,

Midrange = (Highest + Lowest)/2

Midrange = (90 + 78)/2

Midrange = 84

e) Range is the difference the highest and the lowest numbers in a dataset.

Range = 90 - 78 = 12

f) Variance is an average of the squared deviations from the mean.

Mathematically,

Variance = [Σ(x - xbar)²/N]

xbar = mean

Σ(x - xbar)² = (78 - 84.14)² + (80 - 84.14)² + (83 - 84.14)² + (85 - 84.14)² + (86 - 84.14)² + (87 - 84.14)² + (90 - 84.14)² = 102.8572

Variance = (102.8572)/7

Variance = 14.69

g) Standard deviation = √(variance)

Standard deviation = √(14.69)

Standard deviation = 3.83

Hope this Helps!!!

8 0

4 years ago

Read 2 more answers

What is the difference between a circuit and a current?

dimulka [17.4K]

Answer:

a circuit is a complete loop that carries a current in the form of electrons from negative to positive

Explanation:

The difference between current and circuit is that a circuit is a complete loop that carries a current in the form of electrons from negative to positive .

In a circuit, there are difference elements or components such as the battery, wire, resistor. The goal of a circuit arrangement is to completely carry current from one end to another.

The current is the quantity of charge that flows within the circuit per unit of time.

So, the battery supplies the electromotive force to move the current round the circuit.

3 0

3 years ago

Concave lenses have

Scilla [17]

Answer:

A. negative; virtual.

Explanation:

A concave lens always forms a virtual erect and diminished image. The reason why is that the image is actually formed by the intersection of virtually extended refracted rays.

Hope this helps!

Please mark as brainliest if correct!

3 0

2 years ago

Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou

kirza4 [7]

Answer:

(a) 43.2 kC

(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

i = current flow = 3 A
t = time interval for which the current flow = $4\ h = 4\times 3600\ s = 14400\ s$
V = terminal voltage of the battery
R = rate of energy = 9 cents/kWh

<u>Assume:</u>

Q = charge transported as a result of charging
E = energy expended
C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

$\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\$

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

$E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\$

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

$\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents$

Hence, 0.108V cents is the charging cost of the battery.

4 0

4 years ago