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2 years ago

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challenge: A 3 kg model airplane is traveling at a speed of 33 m/s.. The operator then increases the speed up to 45 m/s in 2 sec

onds. How much force did the engine need in order to make this change?

1 answer:

devlian [24]2 years ago

6 0

The average force exerted on an object is given by the equation:

F = Δp/Δt = mΔv/Δt = m(Vf-Vi)/Δt

F is the average force, m is the object's mass, Vi is the object's initial velocity, Vf is the object's final velocity, and Δt is the time elapsed.

Given values:

m = 3kg

Vi = 33m/s

Vf = 45m/s

Δt = 2s

Plug in the values and solve for F:

F = 3(45-33)/2

F = 18N

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A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C

Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 $\frac{J}{kg k}$

The heat of vaporization is = 2256 $\frac{KJ}{kg}$

Mass = 0.030 kg

Energy needed to heat water to boiling point = $m c ( T_{2} - T_{1} )$

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy ( $E_{1}$ ) = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

$E_{2}$ = 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E = $E_{1}$ + $E_{2}$

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

9 0

2 years ago

Sasha is ordered Ampicillin 50mg/kg/day x 48 hours, to be given every 6 hours in 100mls of N/S run over 30 minutes. The tubing h

Anna007 [38]

Answer:

The correct dose = 1454.54 mg

and The jnfusion rate = 41.67 gitt/hr

Explanation: the correct dose will be 50mg/kg × kg/2.2 × 64lb

= 1454.54 mg

infusion rate will be

10 gtts/ml × 50mg/6 × 30/60

Infusion rate = 15000/360

= 41.67 gitt/hr

5 0

2 years ago

Please help me with questions 1, 2 and 3. <br> i need a step by step explanation

kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) $x=\frac{1}{2}at^2+v_0t+x_0$

The equation for velocity v and a constant acceleration a is:

(ii) $v=at+v_0$

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) $a = \frac{v -v_0}{t}$

(iv) $x = \frac{v-v_0}{2t}t^2+v_0t + x_0$

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) $x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t$

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v: $v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}$

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

$f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}$

5 0

2 years ago

A 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what i

never [62]

Answer:

The recoil velocity is 0.2687 m/s.

Explanation:

∵ The person is sitting on an icy surface , we can assume that the surface is frictionless.

∴ There is no force acting acting on the person and book as a system in horizontal direction.

Hence , momentum is conserved for this system in horizontal direction of motion.

If 'i' and 'f' be the initial and final states of this system , then by principle of conservation of momentum(p) -

$p_{i}$ = $p_{f}$

System initially is at rest

∴ $p_{i}$ =0

∴ From the above 2 equations

$p_{f}$ =0

We know that ,

Momentum(p)=Mass of the body(m)×velocity of the body(v)

Let $m_{1}$ and $m_{2}$ be the mass of the person and the book respectively and $v_{1}$ and $v_{2}$ be the final velocities of the person and book respectively.

∴ $p_{f}$ = $m_{1}$ $v_{1}$ + $m_{2}$ $v_{2}$ =0

From the question ,

$m_{1}$ = 74.9 kg

$m_{2}$ = 2.44 kg

$v_{2}$ = 8.25 m/s

Substituting these values in the above equation we get ,

(74.9 × $v_{1}$ )+ (2.44×8.25) = 0

∴ $v_{1}$ = - 0.2687 m/s (Negative sign suggests that the motion of the person is opposite to that of the book)

∴ The recoil velocity is 0.2687 m/s.

4 0

2 years ago

A small sphere with a charge of − 0.60 μc is placed in a uniform electric field of magnitude 1.2 × 106 n/c pointing to the west.

romanna [79]

Force on a charge placed in electric field is given by

$F = qE$

here $q = -0.60 /mu C$

$E = 1.2 * 10^6 N/c$

magnitude of force is given by

$F = 1.2 * 10^6* 0.60 * 10^{-6}$

$F = 0.72 N$

Direction of force is opposite to electric field direction as it is a negative charge

so direction is towards EAST

3 0

2 years ago