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2 years ago

6

Per single gallon of gas, Gina's vehicle can go 16 more miles than Amanda's vehicle. If the combined distance the vehicles can t

ravel on one gallon of gas is 72 miles, what is the distance that Gina's vehicle travels?
A. 24 miles

B. 28 miles

C. 36 miles

D. 44 miles

1 answer:

77julia77 [94]2 years ago

8 0

The answer is D. 44 miles

That's because if we take one vehicle to be X, and another to be X + 16, then to calculate X + Y = 72 we can write it down as <span>x + (x+16)=<span>72

Then we have 2x + 16 = 72
2x = 56
x = 28

The result is then D because </span></span>
28 + 16 = 44.

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The velocity of a 1.3 kg block sliding down a frictionless inclined plane is found to be 1.26 m/s. 1.10 s later, it has a veloci

nasty-shy [4]

Answer:

$\theta = 25.3^\circ$

Explanation:

The acceleration of the block can be found by the kinematics equations:

$v = v_0 + at\\5.88 = 1.26 + a(1.1)\\a = 4.2~m/s^2$

Since the plane is frictionless, the only force acting on the block along the motion of the block is its weight.

$F = mg\sin(\theta) = ma\\g\sin(\theta) = a\\(9.8)\sin(\theta) = 4.2\\\theta = 25.3^\circ$

7 0

3 years ago

Through what vertical height is a 50 N object moved if 250 J of work is done lifting it against the gravitational field of Earth

zimovet [89]

5m

Explanation:

Given parameters:

Weight of object = 50N

Work done in lifting object = 250J

Unknown:

Vertical height = ?

Solution:

The work done on an object is the force applied to lift a body in a specific direction.

Work done = force x distance

Weight is a force in the presence of gravity;

Work done = weight x height of lifting

Height of lifting = $\frac{work done }{weight}$

Height of lifting = $\frac{250}{50}$ = 5m

The vertical height through which the object was lifted is 5m

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6 0

3 years ago

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

$\omega=-\alpha t$

$\alpha=-\dfrac{\omega}{t}$

$\alpha=-\dfrac{50.0}{20.0}$

$\alpha=-2.5\ rad/s^2$

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

$\theta=\omega_{0}t+\dfrac{1}{2}\alpha t$

Put the value into the formula

$\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2$

$\theta=500\ rad$

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

$\vec{\tau}=\vec{r}\times\vec{f}$

$\tau=r\times f\sin\theta$

Put the value into the formula

$\tau=2.5\times4\times 9.8\sin60$

$\tau=84.87\ N-m$

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0

3 years ago

Please please help I'm stuck!!!

Korvikt [17]

Answer: iDc im in 3grade

Explanation:sorry

6 0

2 years ago

Read 2 more answers

g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts,

Stolb23 [73]

Answer:

$523269.9\ \text{N/m}$

Explanation:

q = Charge

r = Distance

$q_1=25\ \text{C}$

$r_1=3000\ \text{m}$

$q_2=40\ \text{C}$

$r_2=850\ \text{m}$

The electric field is given by

$E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}$

The electric field at the aircraft is $523269.9\ \text{N/m}$

4 0

3 years ago