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LenKa [72]
3 years ago
8

an object of mass 50 kg is accelerated uniformly from a velocity of 10m/s to 72km/h in 10 s calculate the initial and final mome

ntum of the object also find the magnitude of the force exerted on the object​
Physics
1 answer:
olganol [36]3 years ago
8 0

<em>hope </em><em>this</em><em> answer</em><em> helps</em><em> you</em><em> </em><em>dear.</em><em>.</em><em>.</em><em>take </em><em>care </em><em>and </em><em> </em><em>may </em><em>u </em><em>have </em><em>a </em><em>great </em><em>day </em><em>ahead</em><em>!</em>

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Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav
Vikki [24]

Answer:

The force constant is  k =1.316 *10^{7} \  N/m

The energy stored in the spring is  E =  1.68 *10^{7} \ J

Explanation:

From the question we are told that

   The mass of the object is  M  = 4.8*10^{5} \ kg

    The period is T  = 1.2 \ s

The period of the spring oscillation is  mathematically represented as

         T  =2 \pi \sqrt{ \frac{M}{k}}

where  k is the force constant

   So making k the subject

       k = \frac{4 \pi ^2 M }{T^2}

substituting values

       k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}

      k =1.316 *10^{7} \  N/m

The energy stored in the spring is mathematically represented  as

       E =  \frac{1}{2} k x^2

Where x is the spring displacement which is given as

        x =  1.6 \ m

substituting values

      E =  \frac{1}{2} (1.316 *10^{7}) (1.6)^2

       E =  1.68 *10^{7} \ J

   

7 0
3 years ago
Given the 1-m stick shown below, which is held by a thread at its center. Block 1 is 15 N held at the 10 cm mark, while block 2
sattari [20]
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7 0
2 years ago
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mrs_skeptik [129]

Answer:

25 to the right

Explanation:

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4 0
3 years ago
What is power?
Blizzard [7]

Answer:

D

Explanation:

P=Work/Time

The rate at which work is done matches that.

6 0
3 years ago
A student pushes on a 20.0 kg box with a force of 50 N at an angle of 30° below the horizontal. The box accelerates at a rate of
PtichkaEL [24]

Answer:

225 N

Explanation:

"Below the horizontal" means he's pushing down at an angle.

Draw a free body diagram of the box.  There are three forces: normal force N pushing up, weight force mg pulling down, and the applied force F at an angle θ.

Sum of forces in the y direction:

∑F = ma

N − mg − F sin θ = 0

N = F sin θ + mg

Plug in values:

N = (50 N) (sin 30°) + (20.0 kg) (10 m/s²)

N = 225 N

8 0
3 years ago
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