PLEASE HELP ANSWER THIS QUESTION !

A 1.70 kg block slides on a horizontal, frictionless surface until it encounters a spring with a force constant of 955 N/m. The

Mariana [72]

Answer:

The initial speed of the block is 1.09 m/s

Explanation:

Given;

mass of block, m = 1.7 kg

force constant of the spring, k = 955 N/m

compression of the spring, x = 4.6 cm = 0.046 m

From principle of conservation of energy

kinetic energy of the block = elastic potential energy of the spring

¹/₂mv² = ¹/₂kx²

mv² = kx²

$v = \sqrt{\frac{kx^2}{m} }$

where;

v is the initial speed of the block

x is the compression of the spring

$v = \sqrt{\frac{955*(0.046)^2}{1.7} } \\\\v = 1.09 \ m/s$

Therefore, the initial speed of the block is 1.09 m/s

4 0

3 years ago

Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are

Shalnov [3]

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm and v is ∞

so we substitute

1/f = 1/(-200 cm) + 1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

4 0

3 years ago

HELPP !! pleaseeeeeeeee!!!!!!!!

sertanlavr [38]

Answer:

Atomic name is your answer.

5 0

3 years ago

Read 2 more answers

. Write the following in the scientific notation of 0.0000002400m

aleksandr82 [10.1K]

Explanation:

<h2>Steps :</h2>

Move decimal from left to right =0 000000240.0
Then count the numbers before decimal and write it like this =240.0x10 power-9
That's all

hope it help

7 0

3 years ago

4.00 kg glass coffee cup is 30.0°C at room temperature. It is then plunged into hot dishwater at a temperature of 90.0°C, as sho

statuscvo [17]

Answer:

Q = 200800 Joules.

Explanation:

Given the following data;

Mass = 4kg

Initial temperature = 30.0°C

Final temperature = 90.0°C

Specific heat capacity of glass = 837 J/kg°C

To find the quantity of heat absorbed;

Heat capacity is given by the formula;

$Q = mcdt$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

dt = T2 - T1

dt = 90 - 30

dt = 60°C

Substituting the values into the equation, we have;

$Q = 4*837*60$

Q = 200800 Joules.

Therefore, the amount of heat absorbed is 200800 Joules.

8 0

3 years ago