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3 years ago

6

How many hydrogen (H) atoms are in one molecule of sulfuric acid, H2SO4? (1 points) 1 2 4 7

2 answers:

NeTakaya3 years ago

8 0

Answer:

2

Explanation:

Vesna [10]3 years ago

4 0

Answer:

Explanation:H2 which is mean 2 atoms of hydrogen

You might be interested in

Which most likely could cause acceleration?

Anastaziya [24]

Answer:

this may seem wierd but a hill

Explanation:

a hill if its steep enough and you have wings on a car or something could make you fly depending the weight wind would also help a stop light no a flat road could if you going fast enough wat movement is depending thiers your 4 options hopee you learned something

8 0

3 years ago

An explosion occurs 34 km away. the time it takes for its sound to reach your ears, traveling at 340 m/s, is

Murrr4er [49]

Let's convert the distance in meters first:
$d=34 km=34000 m=3.4 \cdot 10^4 m$
The sound wave travels at speed
$v=340 m/s$
So we can use the basic relationship between space, time and velocity to calculate the time it takes for the sound to reach our ears:
$t= \frac{S}{v}= \frac{3.4 \cdot 10^4 m}{340 m/s}=100 s$

3 0

3 years ago

A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t

MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2 m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0

3 years ago

What is the kinetic energy of a 3.8kg ball that is rolling at 2.2 m/s

labwork [276]

Kinetic energy = 0.5 x m x v²
Kinetic energy = 0.5 x 3.8 x 2.2²
Kinetic energy = 9.196 J

3 0

3 years ago

A .183 kg ball is moving 18.8 m/s when it runs into a spring of spring constant 86.9 N/m. How much KE does the ball have when it

Lemur [1.5K]

Answer:

The value is $KE_B = 20.59 \ J$

Explanation:

From the question we are told that

The mass of the ball is $m = 183 \ kg$

The initial speed of the ball is $u = 18.8 \ m/s$

The spring constant is $k = 86.9 \ N/m$

The compression distance is $x = 0.520 \ m$

Generally the energy stored in the string is mathematically represented as

$E = \frac{1}{2} * k * x^2$

=> $E = \frac{1}{2} * 86.9 * 0.520 ^2$

=> $E = 11.75 \ J$

Generally the kinetic energy of the ball is mathematically represented as

$KE_b = \frac{1}{2} * m * u^2$

=> $KE_b = \frac{1}{2} * 0.183 * (18.8 )^2$

$KE_b = 32.34 \ J$

Generally the KE the ball have when it has compressed the spring is mathematically represented as

$KE_B = KE_b - E$

=> $KE_B = 32.34 - 11.75$

=> $KE_B = 20.59 \ J$

7 0

3 years ago