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alekssr [168]
3 years ago
6

A 4kg object is at rest. How much force is required to get the object to a velocity of 20m/s in 2 seconds? (Show work and includ

e units)
Physics
1 answer:
Ugo [173]3 years ago
8 0

Explanation:

<h2>Answer:-</h2>

The Object was at rest. So, Initial Velocity is Zero.

  • [Initial Velocity]u = 0
  • [Final Velocity]v = 20 m/s
  • [Time]t = 2 seconds
  • Mass = 4kg
  • Force = ?

We know that:-

\sf{Force = Mass \times \dfrac{(v-u)}{t}}

Applying it, we get:-

\sf{Force = 4 \times \dfrac{(20-0)}{2}}

\sf{Force = 2 \times \dfrac{(20)}{2}}

\sf{Force = 2 \times 10}

\sf{Force = 20 \ N \ (Newton)}

Hope it helps :)

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A golf club consists of a shaft connected to a club head. The golf club can be modeled as a uniform rod of length ℓ and mass m1
slega [8]

Answer:

 m₁ (R +ℓ/2) / /( m₁ + m₂ )

Explanation:

We shall consider centre of the club's head  as origin .

The centre of mass of the rod attached with sphere will be away from the centre of sphere by a distance

= R + ℓ /2 and its mass is m₁

mass of sphere is m₂ and distance of its centre of mass from origin is zero

So required CM = m₁ x₁ + m₂ x₂ / ( m₁ + m₂ )

= m₁ (R + ℓ /2) + m₂ x 0 /( m₁ + m₂ )

=  m₁ (R +ℓ/2) / /( m₁ + m₂ )

5 0
3 years ago
A typical reaction time to get your foot on the brake in your car is 0.2 second. If you are traveling at a speed of 60 mph (88 f
katrin [286]

Explanation:

Not enough information. It really depends on the technical details of the car ( the data provided is offering just the human factor of the reaction, not the time for getting the impulse through when using the breaks

5 0
3 years ago
Read 2 more answers
. On a safari, a team of naturalists sets out toward a research station located 9.6 km away in a direction 42° north of east. Af
Elden [556K]

Answer:\theta =49.76^{\circ} North of east

Explanation:

Given

Research station is 9.6 km away in 42^{\circ}North of east

after travelling 3.1 km 25^{\circ} north of east

Position vector of safari after 3.1 km is

r_2=3.1cos25\hat{i}+3.1sin25\hat{j}

Position vector if had traveled correctly is

r_0=9.6cos42\hat{i}+9.6sin42\hat{j}

Now applying triangle law  of vector addition we can get the required vector(r_1)

r_1+r_2=r_0

r_1=(9.6cos42-3.1cos25)\hat{i}+(9.6sin42-3.1sin25)\hat{j}

r_1=4.325\hat{i}+5.112\hat{j}

Direction is given by

tan\theta =\frac{y}{x}=\frac{5.112}{4.325}

\theta =49.76^{\circ}

8 0
3 years ago
The previous part could be done without using the decay equation, because the ratio of original 14C14C to present 14C14C was an
salantis [7]

Answer:

The decay constant is 1.21×10^-4/year

Explanation:

Decay constant = 0.693/half-life

Half-life = 5730 years

Decay constant = 0.693/5730 years = 1.21×10^-4/year

4 0
4 years ago
1. Calculate the power it takes for a 583 N student to climb up1.5 m stairs in 2.1 seconds. (find work first!)GivenFormulaSubsti
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Given data

*The given force is F = 583 N

*The given distance is s = 1.5 m

*The given time is t = 2.1 s

The formula for the power it takes for a 583 N student to climb up 1.5 m stairs in 2.1 seconds is given as

P=\frac{F\mathrm{}s}{t}

Substitute the known values in the above expression as

\begin{gathered} P=\frac{(583)(1.5)}{(2.1)} \\ =416.42\text{ W} \end{gathered}

Hence, the power it takes for a 583 N student to climb up 1.5 m stairs in 2.1 seconds is P = 416.42 W

3 0
2 years ago
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