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andrew11 [14]
3 years ago
7

A ball is thrown horizontally from the top of

Physics
2 answers:
Serhud [2]3 years ago
5 0

Answer:

52.38 m/s

Explanation:

Masja [62]3 years ago
5 0

Answer:

54.23 m/s

Explanation:

-- time to fall from 140m = 5.35 s

-- vertical speed after 5.35 s =

Speed = (5.35) x (9.8) = 52.38 m/s

-- horizontal speed = (75m/5.35s)

Speed = 14.03 m/s

Resultant speed due to vertical & horizontal speed components:

Speed = sqrt(52.38^2 + 14.03^2)

Speed = 54.23 m/s

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A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

7 0
3 years ago
Consider the following arrangement with a frictionless/massless pulley. Determine the force F required to move block A if the co
Delicious77 [7]

Answer: The free - body diagrams for blocks A and B. frictionless surface by a constant horizontal force F = 100 N. Find the tension in the cord between the 5 kg and 10 kg blocks. The string that attaches it to the block of mass M2 passes over a frictionless pulley of negligible mass. The coefficient of kinetic friction Hk between M.

Explanation: Hope this helped :)

6 0
3 years ago
: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma
e-lub [12.9K]

Answer:

x_1 = 3.74m

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

a_s = \dfrac{F}{m}

a_s = \dfrac{8.2}{12}

a_s = 0.68\ m/s^2

F = m a_m

a_m = \dfrac{F}{m}

a_m = \dfrac{8.2}{70}

a_m = 0.12\ m/s^2

x_1+x_2 = 25

\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25

(a_c+a_m)t^2=50

(0.12+0.68)t^2=50

t = \sqrt{\dfrac{50}{0.8}}

t = 7.90 s

x_1 = \dfrac{1}{2}a_ct^2

x_1 = 0.5\times 0.12 \times 7.90^2

x_1 = 3.74m

5 0
3 years ago
write an essay describing and predicting the effects of fitness-related stress management techniques on the body.
Tatiana [17]
This is something u are going to have to do
7 0
3 years ago
LAB: Thermal Engery Transfer
hodyreva [135]

Answer:

A- mass and type of material

B- type of material

C- Temperature

Explanation:

thx

6 0
2 years ago
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