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3 years ago

You drive 3 hours at 50 mph and then 11 hours at 70 mph. What is your average velocity in miles per hour?

Physics

1 answer:

Archy [21]3 years ago

6 0

The average speed of the car is 65.71 mph.

The given parameters;

initial speed of the car, u = 50 mph

time of motion, t = 3 hours

final speed of the motion, v = 70 mph

time of motion, t = 11 hours

The average speed of the car is calculated as follows;

Average speed = total distance divided by the total time of motion.

The total distance = (50 x 3) + (70 x 11) = 920 miles

The total time of motion, = (3 + 11) = 14 hours.

The average speed of the car is;

$average \ speed = \frac{total \ distance}{total \ time } \\\\average \ speed = \frac{920 \ miles }{14 \ hours} \\\\average \ speed = 65.71 \ mph$

Thus, the average speed of the car is 65.71 mph.

Learn more here: the average speed of the car is 65.71 mph.

Learn more here: brainly.com/question/17289046

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The electrons of carbon and hydrogen are _____________ in this molecule of methane.

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Answer:

The answer is C

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3 years ago

A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i

mars1129 [50]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The electric field at that point is $E = 7500 \ N/C$

Explanation:

From the question we are told that

The radius of the inner circle is $r_i = 0.80 \ m$

The radius of the outer circle is $r_o = 1.20 \ m$

The charge on the spherical shell $q_n = -500nC = -500*10^{-9} \ C$

The magnitude of the point charge at the center is $q_c = + 300 nC = + 300 * 10^{-9} \ C$

The position we are considering is x = 0.60 m from the center

Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

$E = \frac{k * q_c }{x^2}$

substituting values

$E = \frac{k * q_c }{x^2}$

where k is the coulomb constant with value $k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

substituting values

$E = \frac{9*10^9 * 300 *10^{-9}}{0.6^2}$

$E = 7500 \ N/C$

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3 years ago

a 77 N net force is applied to a box which slides horizontally across a floor for 6.7 m. What amount of work is done on the box

ArbitrLikvidat [17]

W=fd
W=77*6.7
W=515.9 J

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3 years ago

A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo

m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 . t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

p₀ = m1 v₀₁

Moment after shock

$p_{f}$ = (m1 + m2) $v_{f}$

p₀ = $p_{f}$

m1 v₀₁ = (m1 + m2) $v_{f}$

$v_{f}$ = v₀₁ m1 / (m1 + m2)

$v_{f}$ = 4.4 600 / (600 + 500)

$v_{f}$ = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

Em₀ = K = ½ (m1 + m2) $v_{f}$ ²

After compressing the spring

$E_{mf}$ = Ke = ½ k x²

As there is no rubbing the energy is conserved

Em₀ = $E_{mf}$

½ (m1 + m2) $v_{f}$ ² = = ½ k x²

x = $v_{f}$ √ (k / (m1 + m2))

x = 2.4 √ (11/3000)

x = 0.396 m

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4 years ago

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Explanation:

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3 years ago