Three 3-ohm resistors are connected in parallel. The total resistance for the resistors is

The sensor in the torso of a crash test dummy records the magnitude and direction of the net force acting on the dummy.If the du

Sunny_sXe [5.5K]

Here in crash test the two forces are acting on the dummy in two different directions

As we know that force is a vector quantity so we need to use vector addition laws in order to find the resultant force on it.

So here two forces are given in perpendicular direction with each other so as per vector addition law we need to use Pythagoras theorem to find the resultant of two vectors

so we can say

$F_{net} = \sqrt{F_1^2 + F_2^2}$

here given that

$F_1 = 130.0 N$

$F_2 = 4500.0 N$

now we will plug in all data in the above equation

$F_{net} = \sqrt{4500^2 + 130^}$

$F_{net} = 4501.9 N$

so it will have net force 4501.9 N which will be reported by sensor

4 0

3 years ago

Read 2 more answers

A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___

34kurt

Answer:

The final displacement from the origin is 1.6 km to the NE

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8· $\hat j$ , in vector form

2) 1.0 km East = 1.0· $\hat i$ , in vector form

3) 1.6 km South = -1.6· $\hat j$ , in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8· $\hat j$ + 1.0· $\hat i$ +(-1.6· $\hat j$ ) = 1.2· $\hat j$ + 1.0· $\hat i$

Final displacement, = 1.0· $\hat i$ + 1.2· $\hat j$

Where;

Δx = The displacement in the x-direction = 1.0· $\hat i$

Δy = The displacement in the y-direction = 1.2· $\hat j$

The magnitude of the resultant displacement vector is given as follows

$\left | d \right |$ = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0

3 years ago

A total Δν of 15 km/s is required to achieve an interplanetary mission. The proposed rocket has two stages. The first stage alon

AlexFokin [52]

Answer:

102000 kg

Explanation:

Given:

A total Δν = 15 km/s

first stage mass = 1000 tonnes

specific impulse of liquid rocket = 300 s

Mass flow rate of liquid fuel = 1500 kg/s

specific impulse of solid fuel = 250 s

Mass flow of solid fuel = 200 kg/s

First stage burn time = 1 minute = 1 × 60 seconds = 60 seconds

Now,

Mass flow of liquid fuel in 1 minute = Mass flow rate × Burn time

or

Mass flow of liquid fuel in 1 minute = 1500 × 60 = 90000 kg

Also,

Mass flow of solid fuel in 1 minute = Mass flow rate × Burn time

or

Mass flow of solid fuel in 1 minute = 200 × 60 = 12000 kg

Therefore,

The total jettisoned mass flow of the fuel in first stage

= 90000 kg + 12000 kg

= 102000 kg

3 0

3 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

3 years ago

What is the formula that represents the simplest ratio of the atoms in the components

Dafna1 [17]

Answer:

Empirical formula

Hope it'll help!

stay safe:)

5 0

3 years ago