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dolphi86 [110]
4 years ago
14

A car travels 60.0 miles per hour. What is its velocity in m/s? (1 km = 0.621 mi). . A] 0.01 m/s. .B] 0.03 m/s. .C] 9.66 m/s. .

D] 26.8 m/s
Physics
2 answers:
muminat4 years ago
7 0
By converting miles in to meter 
60 miles=96560 meter
1 hour = 3600 seconds
velocity= 96560m/3600s
Velocity= <span>26.8 m/s</span>
guapka [62]4 years ago
7 0

Answer:

v = 26.8 m/s

so the speed in SI unit will be 26.8 m/s

Explanation:

speed of the car is given as

v = 60 miles/hour

so we will have

1 miles = 1609 meter

1 hour = 3600 s

now we know that

v = 60 (\frac{miles}{hour}) \times (\frac{1609 m}{1 mile})\times (\frac{1 hour}{3600 s})

v = 26.8 m/s

so the speed in SI unit will be 26.8 m/s

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A 3.20 g sample of a salt dissolves in 9.10 g of water to give a saturated solution at 25°C. a. What is the solubility (in g sal
muminat

Answer:

  • The solubility of the salt is 35.16 (g/100 g of water).
  • It would take 71.09 grams of water to dissolve 25 grams of salt.
  • The percentage of salt that dissolves is 52.7 %

Explanation:

<h3>a.</h3>

We know that 3.20 grams of salt in 9.10 grams of water gives us a saturated solution at 25°C. To find how many grams of salt will gives us a saturated solution in 100 grams of water at the same temperature, we can use the rule of three.

\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{100 \ g \ water}

Working it a little this gives us :

x = 100 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water}

x = 35.16 \ g \ salt

So, the solubility of the salt is 35.16 (g/100 g of water).

<h3>b.</h3>

Using the rule of three, we got:

\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{25 \ g \ salt}{x \ g \ water}

Working it a little this gives us :

x = \frac{25 \ g \ salt}{ \frac{3.20 \g \ salt}{9.10 \ g \ water}}

x = 71.09 g \ water

So, it would take 71.09 grams of water to dissolve 25 grams of salt.

<h3>C.</h3>

Using the rule of three, we got that for 15.0 grams of water the salt dissolved will be:

\frac{3.20 \g \ salt}{9.10 \ g \ water} = \frac{x \ g \ salt}{15.0 \ g \ water}

Working it a little this gives us :

x = 15.0 \ g \ water * \frac{3.20 \g \ salt}{9.10 \ g \ water}

x = 5.27\ g \ salt

This is the salt dissolved

The percentage of salt dissolved is:

percentage \ salt \ dissolved = 100 \% * \frac{g \ salt \ dissolved}{g \ salt}

percentage \ salt \ dissolved = 100 \% * \frac{ 5.27\ g \ salt }{ 10.0 \ g \ salt}

percentage \ salt \ dissolved = 52.7 \%

3 0
4 years ago
An electric heater is rated at 1400 W, a toaster is rated at 1150 W, and an electric grill is rated at 1560 W. The three applian
gayaneshka [121]

Answer:

The current in the heater is 12.5 A

Explanation:

It is given that,

Power of electric heater, P₁ = 1400 W

Power of toaster, P₂ = 1150 W

Power of electric grill, P₃ = 1560 W

All three appliances are connected in parallel across a 112 V emf source. We need to find the current in the heater. We know that in parallel combination of resistors the current flowing in every branch of resistor divides while the voltage is same.

Electric power, P_1=V\times I_1

I_1=\dfrac{P_1}{V}

I_1=\dfrac{1400\ W}{112\ V}

I_1=12.5\ A

So, the current in the heater is 12.5 A. Hence, this is the required solution.  

8 0
4 years ago
A 55kg cart is pushed by a force of 225 n. what is the carts acceleration?
gayaneshka [121]
F=ma so a=F/m
a=225/55=4.09 m/sec^2
3 0
4 years ago
The further the planets is from the sun the is its year
Genrish500 [490]
The further the planet is from the sun the smaller the year is

7 0
4 years ago
Stairway must have uniform riser height and tread depth; variations in riser height or tread depth shall not be over _______ inc
alexandr1967 [171]

Answer:

\frac{3}{8} inches

Explanation:

the variations in riser height or tread depth should not be grater than \frac{3}{8} inches that is equal to 9.5 mm but the maximum riser height should be the  \frac{81}{4} inch  but variation in riser height should not exceed to \frac{3}{8} inches. The minimum riser height should be 7 inches which is equal to the 178 mm

5 0
4 years ago
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