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3 years ago

A car travels 30 km north in 25 min. and 40 km east in 35 min. What is the total time in hours? Be careful to carry over

Physics

1 answer:

antoniya [11.8K]3 years ago

7 0

80 minutes when you covert it, it’s 1 hour 20 minutes

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What is your basal metabolism

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It's the energy your body spends to just keep you breathing and your heart beating ... just being alive, without trying to DO anything.

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3 years ago

Can someone pls help me?!!!! i need it asap!! i’m very stressed

Mariulka [41]

Answer:

Acceleration

Explanation:

Its speed or velocity change

6 0

3 years ago

A. State whether the acceleration is positive, negative or zero for each section in the speed versus

bazaltina [42]

Answer:

a=positive

b=0

c=positive

d=negative

Explanation:

a=acceleration depends on the speed and time. if the speed and time are increasing at the same rate, the acceleration value will be positive as the vehicle is speeding up.

b=the speed and time are not increasing, therefore the vehicle is either stationary or travelling at a steady pace.

c=same explanation as a

d=the speed and time are not increasing at the same rate as the speed is decreasing. this means that the car is slowing down

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2 years ago

3. Force = _____ x _____?

harkovskaia [24]

Answer:

Explanation:

Force = mass * acceleration.

8 0

2 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago