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2 years ago

8

A cyclist on a training ride records the distance she travels away from home. The data only shows the first150 minutes of the ri

de before her cycling computer ran out of battery.

1 answer:

Varvara68 [4.7K]2 years ago

3 0

Answer:

A) 58 km

B) 30 mins

Explanation:

In pic details

graph in pic

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You need to repair a gate on the farm. The gate weighs 100 kg and pivots as indicated. A small diagonal bar supports the gate an

tekilochka [14]

Answer:

The force is $F = 3920 \ N$

Explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The weight of the gate is $G = 100\ kg$

The vertical component of F is $F_y = F\ sin \theta$

From the diagram , taking moment about the pivot we have

$W_g * 2 - F_y * 1 = 0$

Where $W_g$ is the weight of the gate evaluated as

$W_g = m_g * g = 100 * 9.8 = 980 \ N$

=> $980 * 2 - Fsin(30) * 1 = 0$

=> $F = \frac{1960}{sin(30)}$

=> $F = 3920 \ N$

7 0

3 years ago

(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the

crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

$E=\frac {-13.6}{n^2}\ eV$

(a) The energy of the electron in n= 6 excited state is:

$E=\frac {-13.6}{6^2}\ eV$

$E=-0.3778\ eV$

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

$E=\frac {-13.6}{1^2}\ eV$

$E=-13.6\ eV$

$Ratio=\frac {E_6}{E_1}$

$Ratio=\frac {-0.3778}{-13.6}$

Ratio = 0.0278

7 0

3 years ago

I.Solve the following problems and answer the following questions. Show all your work and provide answers rounded off to the app

Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

initial velocity of the sprinter, u = 18 km/h

final velocity of the sprinter, v = 27 km/h

time of motion of the sprinter, t = 3.5 x 10⁻⁴ h

Convert the velocity of the sprinter to m/s;

$initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\$

The time of motion is seconds;

$t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s$

The sprinter’s average acceleration is calculated as follows;

$a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2$

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

6 0

2 years ago

Which group contains elements that have the following characteristics:

Harrizon [31]

actually the answer is B because Chlorine, sulfur, and silicon. Chlorine is a halogen and gas. Sulfur forms an ion with a -2 charge in ionic bonds. Silicon is a well-known metalloid.

6 0

2 years ago

Could an experiment similar to young's two-slit experiment be performed with sound? how might this be carried out? does it matte

Arada [10]

Young's double slit experiment(YDSE) can be used for any kind of waves such as electromagnetic waves, sound waves, water waves, gravity waves. YDSE is based on interference. In this experiment, we make two waves interfere in order to obtain bright and dark fringes on the screen(in case of light).
You can carry this out with water, would be great if you try this at pond or water reservoir in order to see perfect ripples.

7 0

3 years ago