All categories

3 years ago

Question 2.

Physics

1 answer:

choli [55]3 years ago

3 0

Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

$1\ \text{inch}=\dfrac{1}{12}\ \text{feet}$

So,

$5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}$

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.

You might be interested in

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago

Which statements correctly describe the properties of air? Select three options. Air has mass but no weight. Air pressure is cre

jolli1 [7]

Answer:

2,4,5

Air pressure is created by the weight of the atmosphere pushing on Earth’s surface.

Denser air is heavier than less dense air.

Air is less dense at higher altitudes.

4 0

3 years ago

Read 2 more answers

I need help with number 6

Otrada [13]

I like your handwriting boy

7 0

3 years ago

Two beams of coherent light are shining on the same piece of white paper. with respect to the crests and troughs of such waves,

defon

Answer:

"where crests and troughs have their maxima at the same time"

Crests and troughs are 180 deg out of phase and when they have their maxima at the same time and place, their net contribution will be zero"

7 0

2 years ago

Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts

olga2289 [7]

Answer:

a) a = - 0.106 m/s^2 (←)

b) T = 12215.1064 N

Explanation:

F₁ = 9*1350 N = 12150 N (→)

F₂ = 9*1365 N = 12285 N (←)

∑Fx = M*a = (M₁ +M₂)*a (→)

F₁ - F₂ = (M₁ +M₂)*a

→ a = (F₁ - F₂) / (M₁ +M₂ ) = (12150-12285)N/(9*68+9*73)Kg

→ a = - 0.106 m/s^2 (←)

(b) What is the tension in the section of rope between the teams?

If we apply ∑Fx = M*a for the team 1

F₁ - T = - M₁*a ⇒ T = F₁ + M₁*a

⇒ T = 12150 N + (9 * 68 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

If we choose the team 2 we get

- F₂ + T = - M₂*a ⇒ T = F₂ - M₂*a

⇒ T = 12285 N - (9 * 73 Kg)*(0.106 m/s^2)

⇒ T = 12215.1064 N

4 0

3 years ago