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Ray Of Light [21]
3 years ago
14

What type of energy is at work when mini-blinds are being raised or lowered

Physics
1 answer:
ohaa [14]3 years ago
4 0
The type of energy that is ar work when blinds are being raised or lowered is potential energy. because potential energy is the energy stored in an object due to its vertical position, the main cause of the potential energy is the acceleration due to gravity. so that is why its formual is PE = mgh
where m is the mass, g is the acceleration due to gravity and h is the height
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Suppose that at a price of $2.60, the quantity of output demanded is 17, and at a price of $6.30, the quantity of output demande
Natalija [7]

Answer: elastic (e = 2.43)

Explanation:

The price elasticity formulae is given below as

Elasticy of price = change in quantity demanded / change in price.

P1 =$2.60, P2 = $6.30, q1 =17 and q2 = 8

e = q2 - q1/ P2 - P1

e = 8 - 17/ 6.30 - 2.60

e = - 9 /3.7

e = - 2.43

We take the modulus of e to have a positive value. Hence e = 2.43

Since e is greater than 1, then the elasticity of demand is elastic

7 0
3 years ago
A drunken sailor stumbles 580 meters north, 530 meters northeast, then 480 meters northwest. What is the total displacement and
kvv77 [185]

Answer:

(a)  1294.66 m

(b) 88.44°

Explanation:

d1 = 580 m North

d2 = 530 m North east

d3 = 480 m North west

(a) Write the displacements in vector forms

\overrightarrow{d_{1}}=580\widehat{j}

\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )

\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}

\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )

\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}

The resultant displacement is given by

\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}

\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}

\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}

magnitude of the displacement

d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m

d = 1294.66 m

(b) Let θ be the angle from + X axis direction in counter clockwise

tan\theta =\frac{1294.18}{35.36}=36.6

θ = 88.44°

4 0
3 years ago
Which of the following means that an image is real?
DochEvi [55]
To declare an image is real you would call it “authentic”
8 0
3 years ago
Read 2 more answers
What is the mass of a neutron?
Anastaziya [24]

Answer:

1

Explanation:

8 0
2 years ago
Read 2 more answers
Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
netineya [11]

Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{2}} and C is a constant

6 0
3 years ago
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