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Readme [11.4K]
3 years ago
7

What force must be provided to accelerate a 64-lb object upward at a rate of 2 ft/s2? (Use g = 32 ft/s2.)

Physics
1 answer:
vovikov84 [41]3 years ago
8 0

Answer:

The force that must be provided to accelerate the object = 302.196 N

Explanation:

<em>Force</em>: Force can be defined as the product of mass and acceleration. The S.I unit of force is Newton (N).

Ft - W = ma............ Equation 1

Ft = W + ma .......... Equation 2

Where Ft = force provided, W = weight of the object, m= mass of the object, a = acceleration of the object.

<em>Given: mass = 64 lb, g = 32 ft/s² a = 2 ft/s²</em>

<em>Conversion: (i)  from 64 pounds to Kg = 64/2.2046</em>

<em>                       = 29.03 kg</em>

<em>   (ii)  from 2 ft/s² to m/s² = 0.3048 × 2 = 0.6098 m/s²</em>

<em> (iii) from 32 ft/s² to m/s² = 0.3048×32 = 9.8 m/s².</em>

and W = mg = 29.03 × 9.8 = 284.494 N,

Substituting these values into equation 2,

Ft = 284.494 + 29.03×0.6098

Ft = 284.494 + 17.7

Ft = 302.196 N

Therefore, the force that must be provided to accelerate the object = 302.196 N

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If wave A has twice the amplitude and three times the frequency as wave B, then the energy carried by wave A must be ____ times
soldier1979 [14.2K]

Answer:

4 times

Explanation:

As we know that the energy of a wave is directly proportional to the square of the amplitude of the wave,

Here, the amplitude of the wave A is twice as compared to B.

So, the energy of wave A is 4 times the energy of wave B.

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4 years ago
A motorcycle is speeding along at a velocity of120kph.Then, it changes to a velocity of 150 kph for 2 minutes.What is the motorc
Taya2010 [7]

Answer:

Explanation:

a=v-u/t

a=acceleration

v=final velocity

u=initial velocity

t=tme taken

we need to convert from kph to ms⁻¹

v= 150*1000/60*60= 41.67ms⁻¹

u= 120*1000/60*60= 33.33ms⁻¹

t= 2*60= 120s

a=41.67-33.33/120

a=8.34/120

a=0.0694ms⁻²

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3 years ago
A natural force of attraction exerted by the earth upon objects, that pulls
Mashutka [201]

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3 years ago
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
3 years ago
How much heat is needed to change the temperature of 3 grams of gold (c = 0.129 ) from 21°C to 363°C? The answer is expressed to
Temka [501]

Q= mcΔT

Where Q is heat or energy

M is mass, c is heat capacitance and t is temperature

You have to convert Celsius into kelvin in order to use this formula I believe

Celsius + 273 = Kelvin

21 + 273 = 294K

363 + 273 = 636K

Now...

Q= (0.003)(0.129)(636-294)

Q= 0.132 J if you are using kilograms, in terms of grams which seems more appropriate the answer would be 132J of energy.  

3 0
3 years ago
Read 2 more answers
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