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3 years ago

15. What is a thermograph?

Physics

2 answers:

Genrish500 [490]3 years ago

8 0

Answer: A device that uses infrared sensors.

Explanation:

Anni [7]3 years ago

8 0

Answer:

a device that uses infrared sensors to create pictures that show variations in temperature

Explanation:

Gradpoint

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In February 1955, a paratrooper fell 365 m from an airplane without being able to open his chute but happened to land in snow, s

DaniilM [7]

Answer:

a. $i=4760 kg*m/s$

b. $D_U= 1.11 m$

Explanation:

F= 120,000N

Kinetic energy @ impact = 120,000*depth

$K_E= (1/2)*85kg*(56m/s)^2$

$K_E=133280 J$

$D_U= \frac{133280J}{120000N} = 1.11 m$

The momentum is equal to the impulse on him from the snow so:

$p=m*v$

$p=85kg*56m/s$

$i=4760 kg*m/s$

7 0

3 years ago

Which answer choice?? I have trouble with this physics concept :/

Zepler [3.9K]

Answer:

Your weight

Explanation:

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3 years ago

List three examples each of objects that are commonly measured by mass by volume and by length

Simora [160]

length- roads, yard stick, square footage in a room

6 0

3 years ago

At which point is the kinetic energy the greatest?

mr_godi [17]

Answer:

I don't know

Explanation:

I think it's at the highest, learned this in seventh grade

6 0

3 years ago

Read 2 more answers

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:

Given:

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

Assume:

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