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Genrish500 [490]
3 years ago
10

An ___ is a deep valley on land that forms along a divergent boundary

Physics
2 answers:
Gennadij [26K]3 years ago
7 0
This is a Rift Valley. 

Hope this helped!
Marta_Voda [28]3 years ago
7 0
An RIFT VALLEY is a deep valley on land that forms along divergent boundary.
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A student is constructing an investigation on static electricity. The student has three balloons and rubs two of them on a piece
Kamila [148]

Answer:

Explanation:

As given, the student has three balloons and rubs two of them on a piece of wool. The rubbing of balloon on wool is the independent variable as it was done on two and not on the third as control.

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2 years ago
How to get stud multipliers in lego star wars the skywalker saga?.
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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
Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w
Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

=  (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0
3 years ago
Students measured the mass of 25.0 mL of water and found it be 25.4 g. The accepted mass is 25.0 g. What is the percent error of
Andre45 [30]
Well first of all, I think the students may have been correct. 
If they didn't use distilled water, and if it wasn't exactly at 
standard temperature, then the mass of  25.0 mL  could
very well be  25.4 grams.  We don't know that there was
any 'error' in their measurement at all.
But the question says there was, so we'll do the math:

The 'error' was  (25.4 - 25.0) = +0.4 gram

As a fraction of the 'real' value, the error was

                            +0.4 / 25.0  =  +0.016 .

To change a decimal to a percent, move the
decimal point two places that way  ===> .

                           + 0.016  =  +1.6 % .

     
Their measurement was 1.6% too high.

Let's not call it an 'error'.  Let's just call it a 'discrepancy'
between the measured value and the 'accepted' value.  OK ?
4 0
3 years ago
Read 2 more answers
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