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Svet_ta [14]
2 years ago
9

A runner starts at position A. He runs 40 m North, 10 m East and 40 m

Physics
1 answer:
Alecsey [184]2 years ago
8 0
The answer is C.) 10 m East
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The treasure map gives the following directions to the buried treasure ​
iren2701 [21]

Answer:

North

South

East

West

Explanation:

please mark me as brainliest

3 0
3 years ago
You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring th
AfilCa [17]

Answer:

\partial \theta = 0.003

Explanation:

we know that

sin\theta = \frac{3.8}{146.4}

\theta = sin^{-1} \frac{3.8}{146.4}

\theta = 1.484°

\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians

as we see that sin\theta = \theta

relative error\frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}

Where X_1 IS HEIGHT OF ROCK

X_2 IS THE HEIGHT OF ROAD

\partial X = uncertainity in measuring  distance

\partial X = 0.05

Putting all value to get uncertainity in angle

\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}

solving for \partial \theta we get

\partial \theta = 0.003

3 0
3 years ago
a balloon inflated in a room at 297k has a volume of 4.00 l. the balloon is then heated to a temperature of 331 k. what is the n
Naddika [18.5K]
V2 = 4.4579 L

Since pressure is constant, use Charle’s law.
Charles's law, a statement that the volume occupied by a fixed amount of gas is directly proportional to its absolute temperature, if the pressure remains constant.

V(olume) 1 = V(olume) 2
————— —————
T(emperature) 1 T(emperature)2

4.00 L = V2
———- ———
297 K 331 K

Cross multiply
(4.00 L x 331 K) = (297 K x V2)
Simplify
1324 L/K = 297 K x V2
Isolate V2 by dividing out 297 K
1324 L/K = V2
————
297 K
(This cancels out the kelvin and leaves you with Liters as the volume measure)

V2 = 4.4579 L

Round to significant digits if required
7 0
2 years ago
En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a
Maru [420]

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

7 0
3 years ago
What is the study of energy and how it affects matter called?
bogdanovich [222]
Physics

I hope I helped:)
6 0
3 years ago
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