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3 years ago

14

What is the deceleration (in m/s2) of a rocket sled if it comes to rest in 1.7 s from a speed of 1020 km/h? (Such deceleration c

aused one test subject to black out and have temporary blindness.)

1 answer:

fgiga [73]3 years ago

6 0

Answer:

40.47m/s²

Explanation:

Acceleration is the change in velocity of a body with respect to time.

Acceleration = change in velocity/time

Given speed = 1020km/hr

Time = 1.7s

Converting 1020km/hr to m/s we have;

(1020×1000/3600)m/s

= 1,020,000/3,600

= 283.3m/s

Acceleration = 283.3/7

Acceleration = 40.47m/s²

Deceleration is therefore 40.47m/s²

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Engineers are trying to improve a race car. Their goal is to increase the acceleration of the car using the same engine. Which c

noname [10]

Decreasing the mass of the car will help increase the acceleration of the car using application of Newton’s laws of motion.

<h3>What is Newton's law of motion?</h3>

This law discusses the relationship between motion of a substance and the forces acting on it.

Heavy objects are more difficult to move which is why decrease in mass will help increase the acceleration.

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2 years ago

The total weight of a car with passengers is 30,000N. The area of contact of each of the four tyres with the ground is 0.025m2.

vlabodo [156]

Answer:

<em>The minimum pressure car tire pressure is 300,000 Pa</em>

Explanation:

<u>Pressure</u>

It's a measure of the force exerted by an object per unit surface:

$\displaystyle P=\frac{F}{A}$

Where F is the force applied on a surface area A.

We are given the total weight of a car with passengers as Ft=30,000 N and we also know the area of contact of each tire with the ground as $A=0.025 m^2$ .

The minimum pressure each tire can have is when the total weight is evenly distributed in all four tires, thus F=Ft/4=7,500 N.

Calculating the pressure:

$\displaystyle P=\frac{7,500}{0.025}$

P=300,000 Pa

The minimum pressure car tire pressure is 300,000 Pa

7 0

3 years ago

A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then t

lapo4ka [179]

Answer:

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

Explanation:

Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ( $P_{atm}$ ), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

$P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}$

Where:

$P_{1}$ , $P_{2}$ - Water total pressures inside the tank and at ground level, measured in pascals.

$\rho$ - Water density, measured in kilograms per cubic meter.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{1}$ , $v_{2}$ - Water speeds inside the tank and at the ground level, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the tank and ground level, measured in meters.

Given that $P_{1} = P_{2} = P_{atm}$ , $\rho = 1000\,\frac{kg}{m^{3}}$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{1} = 0\,\frac{m}{s}$ , $z_{1} = 6.9\,m$ and $z_{2} = 4.9\,m$ , the expression is reduced to this:

$\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)$

And final speed is now calculated after clearing it:

$v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}$

$v_{2} \approx 6.263\,\frac{m}{s}$

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

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4 years ago