Ask question

Ask question

All categories

3 years ago

10

Suppose a child drives a bumper car head on into the side rail, which exerts a force of 3400 N on the car for 0.400 s. (Assume t

he initial velocity is in the positive direction.) (a) What impulse (in kg · m/s) is imparted by this force? (Indicate the direction with the sign of your answer.) kg · m/s (b) Find the final velocity (in m/s) of the bumper car if its initial velocity was 3.50 m/s and the car plus driver have a mass of 200 kg. You may neglect friction between the car and floor. (Indicate the direction with the sign of your answer.) m/s

1 answer:

babunello [35]3 years ago

3 0

Answer:

a) I = -1360 N s b) $v_{f}$ = -3.3 m / s

Explanation:

a) The expression of the impulse is

I = F t = Δp

The force applied by the rail is in the opposite direction to the movement, so it is negative

I = -3400 0.400

I = -1360 N s

b) We use the momentum relationship with the moment

I = ΔP

I = m $v_{f}$ - mv₀

$v_{f}$ = (I + m v₀) / m

$v_{f}$ = (-1360 + 200 3.5) / 200

$v_{f}$ = -3.3 m / s

You might be interested in

An object is thrown vertically and has a speed of 25 m/s when it reaches 1/4 of its maximum height above the ground (assume it s

Crazy boy [7]

Answer:

$v_{i} =28.86\frac{ft}{s}$

Explanation:

Conceptual analysis

We apply the kinematic formula for an object that moves vertically upwards:

$(v_{f} )^{2} =(v_{i} )^{2} -2*g*y$

Where:

$v_{f}$ : final speed in ft/s

$v_{i}$ : initial speed in ft/s

g: acceleration due to gravity in ft/s²

y: vertical position at any time in ft

Known data

For $v_{f} = 25\frac{ft}{s}$ , $y=\frac{1}{4} h$ ; where h is the maximum height

for y=h, $v_{f} =0$

Problem development

We replace $v_{f} = 25\frac{ft}{s}$ , $y=\frac{1}{4} h (ft)$ in the formula (1),

[ $25^{2} =(v_{i} )^{2} -2*g*\frac{h}{4}$ Equation (1)

in maximum height(h): $v_{f} =0$ , Then we replace in formula (1):

$0=(v_{i} )^{2} - 2*g*h$

$2*g*h=(v_{i} )^{2}$

$h=\frac{(v_{i})^{2} }{2g}$ Equation(2)

We replace (h) of Equation(2) in the Equation (1) :

$25^{2} =(v_{i} )^{2} -2g\frac{\frac{(v_{i})^{2} }{2g} }{4}$

$25^{2} =(v_{i} )^{2} -\frac{(v_{i})^{2} }{4}$

$25^{2} =\frac{3}{4} (v_{i} )^{2}$

$v_{i} =\sqrt{\frac{25^{2}*4 }{3} }$

$v_{i} =28.86\frac{ft}{s}$

7 0

4 years ago

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0

4 years ago

Electrons moving back and forth through a circuit would be

yaroslaw [1]

The particles that carry charge through wires in a circuit are mobile electrons. The electric field direction within a circuit is by definition the direction that positive test charges are pushed. Thus, these negatively charged electrons move in the direction opposite the electric field.

5 0

4 years ago

If an object measures to be 80 centimeters in length, then it is equivalent to ____

Lana71 [14]

Answer:

31.496 inches

Explanation:

7 0

4 years ago

Is shelfstone above or below ground

dexar [7]

It believe it is under the ground because it normally is in tunnels

6 0

3 years ago