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a_sh-v [17]
3 years ago
8

Answer them all please

Physics
1 answer:
Amanda [17]3 years ago
5 0

Answer:

respiratory = nose

circulatory = heart

digestive = mouth

nervous = brain

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If a student leaves 100-w lightbulb on In his room for 1 h, how many kilowatt-hour of electricity does the bulb use?
Anettt [7]

1)  0.1 kWh

Explanation:

The power of the lightbulb is:

P = 100 W=0.1 kW

The energy used by the lightbulb is equal to the product between the power (P) and the time (t):

E=Pt

Since the bulb has been left on for t = 1 h, the total electrical energy used is

E=Pt=(0.1 kW)(1 h)=0.1 kWh


2) $ 0.018

Explanation:

The company charges $0.18 per each kilowatt-hour. So, we can find the total cost charged for using 0.1 kWh of electricity by setting the following proportion:

$0.18 : 1 kWh = x : 0.1 kWh

And solving for x, we find

x=\frac{\$0.18 \cdot 0.1kWh}{1 kWh}=\$ 0.018

4 0
3 years ago
Which of the following is the first step in the technological design process?
AVprozaik [17]
B generate ideas not 100 percent positive though
4 0
4 years ago
The foundation of psychology is?<br> Case Studies<br> O Experiments<br> O Research<br> O Analysis
Serggg [28]

Answer:

Case Studies

Explanation:

A case study in psychology is a descriptive research approach used to obtain in-depth information about a person, group, or phenomenon.Case studies use techniques such as personal interviews, direct observation, psychometric tests, and archival records to gather information.

4 0
4 years ago
53. A recently discovered planet has a mass four times as great as Earth's and a radius twice as large as Earth's. What will be
marusya05 [52]

Answer:

g' = g = 9.81 m/s^2

so gravity will be same as that of surface of earth

Explanation:

As we know that the acceleration due to gravity is given as

g = \frac{GM}{R^2}

here we have

M = 4M_e

R = 2R_e

we know that for earth we have

g = 9.81 = \frac{GM_e}{R_e^2}

now if the radius and mass is given as above

g' = \frac{G(4M_e)}{(2R_e)^2}

g' = \frac{GM_e}{R_e^2}

g' = g = 9.81 m/s^2

so gravity will be same as that of surface of earth

6 0
3 years ago
A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta
Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, \theta = 54^{\circ}

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, R = \frac{v^{2}sin2\theta}{g}

R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m

Time of flight, t' = \frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m

Distance traveled vertically:

s_{y} = vcos\theta - \frac{1}{2}gt^{2}

s_{y} = 12.4sin54^{\circ}\times 1.42 -  \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m

Now,

s_{x} = \frac{mx + m'x'}{M}

10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}

x' = 3.535 m

Similarly,

s_{y} = \frac{my + m'y'}{M}

4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m

The location of the other fragment is at (3.535, 1.162)

5 0
4 years ago
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