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2 years ago

A car travels 300 meters in 15 seconds. What is

2 answers:

6 0

C; to get the unit rate or the time it travels per second you divide the total amount traveled by the amount it took to travel. 300/15=20 20•15=300

Katarina [22]2 years ago

5 0

C is what I believe the answer would be

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An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0

2 years ago

a partially inflated weather balloon has a volume of 1.56 * 10^3 L and a pressure of 98.9 lPa. What is the volume of the balloon

e-lub [12.9K]

To solve the problem, we enumerate all the given first. Then the required and lastly the solution.

Given:

V1= 1.56x10^3 L = 1560 L P2 = 44.1 kPa
P1 = 98.9 kPa

Required: V2

Solution:

Assuming the gas is ideal. Ideal gas follows Boyle's Law which states that at a given temperature the product of pressure and volume of a gas is constant. In equation,

PV = k

Applying to the problem, we have

P1*V1 = P2*V2
(98.9 kPa)*(1560 L) = (44.1 kPa)*V2
V2 = 3498.5 L

<em>ANSWER: V2 = 3498.5 L</em>

7 0

2 years ago

What is the magnitude of the force a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away?

sweet [91]

The magnitude of the force<span> a 1.5 x 10-3 C charge exerts on a 3.2 x 10-4 C charge located 1.5 m away is 1920 Newtons. The formula used to solve this problem is:

F = kq1q2/r^2

where:
F = Electric force, Newtons
k = Coulomb's constant, 9x10^9 Nm^2/C^2
q1 = point charge 1, C
q2 = point charge 2, C
r = distance between charges, meters

Using direct substitution, the force F is determined to be 1920 Newtons.</span>

7 0

2 years ago

The diagram shows two vectors that point west and north. What is the magnitude of the resultant vector? 13 miles 17 miles 60 mil

Kipish [7]

Using the formula A squared plus B squared equals C squared, we can find the solution by substituting 5 for A and 12 for B.

By squaring 5, we get 25, and by squaring 12, we get 144. Adding these, we get 169. The square root of this is 13.

6 0

2 years ago

Read 2 more answers

A certain satellite travels in an approximately circular orbit of radius 2.0 × 106 m with a period of 7 h 11 min. Calculate the

kap26 [50]

Answer: Mass of the planet, M= 8.53 x 10^8kg

Explanation:

Given Radius = 2.0 x 106m

Period T = 7h 11m

Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.

This is represented by the equation

T^2 = ( 4π^2/GM) R^3

Where T is the period in seconds

T = (7h x 60m + 11m)(60 sec)

= 25860 sec

G represents the gravitational constant

= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet

Making M the subject of the formula,

M = (4π^2/G)*R^3/T^2

M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2

Therefore Mass of the planet, M= 8.53 x 10^8kg

5 0

2 years ago