What is the net force acting on the buggy?<br> N<br> The net force is pointing to the...

A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about

Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[ $0^{2} -11^{2}$ ]

x-x₀=6.05* $10^{-2}$ m

=6.05 cm

3 0

2 years ago

A Wooden block has a mass of 0.200kg, a specific heat of 710 J (kg times degrees Celsius and is at a temperature of 20.0 degrees

olchik [2.2K]

Answer:

35°C

Explanation:

q = mCΔT

2130 J = (0.200 kg) (710 J/kg/°C) (T − 20.0°C)

T = 35°C

8 0

2 years ago

A weather balloon is rising vertically from a launching pad on the ground. A technician standing 300 feet from the launching pad

avanturin [10]

Answer:

$\dfrac{dh}{dt} =5\ ft/s$

Explanation:

Let

h = height of balloon (in feet).

θ = angle made with line of sight and ground (in radians).

h = 300 tanθ

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

now $\dfrac{dh}{dt}$ can be written as

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{d\theta }{dt} = \dfrac{1}{120}\at \ \theta =\dfrac{\pi}{4}$

When θ = π/4,

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

$\dfrac{dh}{d\theta } = 600$

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{dh}{dt} =600\times \dfrac{1}{120}$

$\dfrac{dh}{dt} =5\ ft/s$

5 0

2 years ago

For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir

abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s

Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s

Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

v = 5.1 m/s

Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s

hence closest is in case C at 5.1 m/s

7 0

2 years ago

Read 2 more answers

Help ASAP). Draw additional masses to show how the block can be made to move toward point A. Be sure to draw the string and labe

chubhunter [2.5K]

Answer:

Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.

4 0

2 years ago

What is the net force acting on the buggy? N The net force is pointing to the...