A car with a mass of 833 kg rounds an unbanked curve in the road at a speed of 28.0 m/s. If the

What are two ways to increase impulse?

Pepsi [2]

Answer:

increase the time the force acts or you could increase the number of temptations.

hope this helped!

5 0

3 years ago

Propose a hypothesis for how the position of the ball will affect the amount of its gravitational pull energy

Ray Of Light [21]

throwing a ball up initially has a lot of kinetic energy because it is moving upwards ( kinetic energy is energy which a body possesses by virtue of being in motion.) this all then get converted to gravitational potential energy, and for a moment it is stationary before it begins to fall again. by the time it has returned again, all the gravitational potential energy has turned back into kinetic.

4 0

3 years ago

where would information on the chemical and physical properties of a specific chemical be located in a laboratory or in the work

goldenfox [79]

Answer:

Both

Explanation:

8 0

3 years ago

A wave with a speed of 9 m/s and a frequency of 0.5 Hz has a λ of what?

patriot [66]

Wave speed = (wavelength) x (frequency)

Wavelength = (wave speed) / (frequency)

Wavelength = (9 m/s) / (0.5 Hz)

Wavelength = 18 m

6 0

3 years ago

A circular sign has a diameter of 40 cm and is subjected to normal winds up to 150 km/h at 10°C and 100 kPa. Determine the drag

Marat540 [252]

Answer:

147.7 N

221.55 Nm

Explanation:

P = Pressure = 100000 Pa

$R_s$ = Mass-specific gas constant = 287.015 J/kg k

T = Temperature = 10+273 = 283 K

C = Drag coefficient = 1.1

A = Area

r = Radius = 0.2 m

v = Speed of wind = $\frac{150}{3.6}\ m/s$

L = Length of pole

Density

$\rho=\frac{P}{R_sT}\\\Rightarrow \rho=\frac{100000}{287.058\times 283}\\\Rightarrow \rho=1.2309\ kg/m^3$

Drag force

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2309\times 1.1\times \left(\pi \times 0.2^2\right)\times \left(\frac{150}{3.6}\right)^2\\\Rightarrow F=147.7\ N$

Force on the circular sign is 147.7 N

$M=F\times L\\\Rightarrow M=147.7\times 1.5\\\Rightarrow M=221.55\ Nm$

Bending moment at the bottom of the pole is 221.55 Nm

6 0

3 years ago