# The engine of an automobile develops 176 hp when rotating at a speed of 1340 rpm. What torque (in foot-pounds) does it deliver?

I tried using some kinematic equations, but that didn't work at all. Then I tried doing something where I would take an integral, but then I don't know how far it went. I searched the website, and everywhere else online, so this is my last resort. Thanks in advance.

## Torque * angular velocity = power

176 hp = 131,243 Watts or N m/s

1340 rpm = 140 radians/s

so Torque = (131,243 / 140) Newton meters

## 937 N m = 691 foot pounds

## To calculate the torque delivered by the engine, you need to know the power developed by the engine and the rotational speed at which it is operating. In this case, you have the power (176 hp) and the rotational speed (1340 rpm), so we can proceed.

First, let's convert the power from horsepower (hp) to watts (W). We'll use the conversion factor: 1 hp = 745.7 W.

176 hp × 745.7 W/hp = 131225.2 W

Next, let's convert the rotational speed from revolutions per minute (rpm) to radians per second (rad/s). We'll use the conversion factor: 1 rpm = 2π/60 rad/s.

1340 rpm × 2π/60 rad/s = 140.52 rad/s

Now, we can calculate the torque (τ) using the formula:

Power (P) = Torque (τ) × Rotational Speed (ω)

τ = P / ω

Substituting the values we have:

τ = 131225.2 W / 140.52 rad/s = 933.59 N·m

Finally, to convert Newton-meters (N·m) to foot-pounds (ft·lb), we can use the conversion factor: 1 N·m = 0.73756 ft·lb.

933.59 N·m × 0.73756 ft·lb/N·m = 688.42 ft·lb

Therefore, the engine delivers a torque of approximately 688.42 foot-pounds.