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3 years ago

Instructions:Select all the correct answers.

2 answers:

7 0

Correct option - 2

As the golf ball hits the ground it exerts some force on the ground.In return the ground will exert an opposite and equal force on the golf ball.

Hope This Helps You!

melomori [17]3 years ago

6 0

The answer would be B: The ground exerts an equal force on the golf ball.

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What is strain?

Sholpan [36]

Answer:

C) The ratio of the change in an object's length to its original length when stretched or compressed.

Explanation:

The formula for strain is:

Strain = Change in Length/Origin Length

Hence, it can be described as the percentile of change in the dimension with respect to the original dimension. So, whenever a tensile or a compressive force is applied on a body, its length changes. The ratio of this change to original length is called strain. So, the correct option is:

C) The ratio of the change in an object's length to its original length when stretched or compressed.

3 0

3 years ago

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0

3 years ago

Which layer of the sun is responsible for producing the light shown in the picture above?

Eva8 [605]

Answer:

C. Photosphere

Explanation:

The lights shown in the figure comes from the outermost layer of the Sun. This layer is called photosphere.

This is the layer from where the light of the Sun is radiated, before travelling through space and reaching us.

The photosphere is the coldest layer of the Sun: its surface temperature is between 4500 and 6000 K. Its width is approximately 100 km.

A characteristic of the photosphere is the presence of the sunspots, which appear as darker spots, and are regions of lower temperature caused by a concentration of magnetic flux.

3 0

3 years ago

Read 2 more answers

A .5kg bird is perched on its nest so that it has 50J of potential energy. how far is it off the of the ground?

pshichka [43]

It is 10.20 m from the ground.

Explanation:

Given:

m = 0.5 kg

PE = 50 J

We know that the Potential energy is calculated by the formula:

$P. E = m \times g \times h$

where m is the is mass in kg; g is acceleration due to gravity which is 9.8 m/s and h is height in meters.

PE is the Potential Energy.

Potential Energy is the amount of energy stored when an object is stationary.

Here, if we substitute the values in the formula, we get

$P. E = m \times g \times h$

50 = 0.5 × 9.8 × h

50 = 4.9 × h

$h = \frac {50} {4.9}$

h = 10.20 m

3 0

3 years ago

The gold has a density of 19300 kg/m3 calculate the mass of one gold bar 1= 2.54cm

icang [17]

Fair enough, but you'll have to tell us the volume of the bar first.

5 0

3 years ago