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k0ka [10]
4 years ago
11

An object falls 210 m in 6.3 s. What was it's initial velocity? Express your answer with the correct number of significant digit

s.
Physics
1 answer:
KIM [24]4 years ago
8 0

Answer:

u = 2.5 \ m/s

Explanation:

Given data:

Height, h = 210 \ m

Time, t = 6.3 \ s

Let the initial velocity of the object be <em>u</em>.

From the kinematic equation,

h = ut + \frac{1}{2}gt^{2}

210 = u \times 6.3 + 0.5 \times 9.80 \times (6.3)^{2}

\Rightarrow \ u = 2.46 \ m/s

u = 2.5 \ m/s (rounding to tenth place)

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I recently did this topic in science class - the answer is A ;)
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3 years ago
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You measure the pressure of the four tires of your car each to be 35.0 pounds per square inch (psi). You then roll your car forw
Oksi-84 [34.3K]

Answer:

Weight of car = 36034.88 lb.ft/s²

Explanation:

We are told the pressure of the four tires of your car is; P = 35.0 psi

Also, the surface area of contact is; A = 32 in²

Thus;

Weight of tires = Pressure × Area

W_tires = 35 × 32

Weight_tires = 1120lb.

To get the weight of the car, we will multiply the tire weight by acceleration due to gravity.

It's value in ft/s² is g = 32.174 ft/s²

Thus;

Weight of car = 1120 × 32.174

Weight of car = 36034.88 lb.ft/s²

5 0
4 years ago
8) Find the X and Y component of 10degree vector that has 5N.
Deffense [45]

Answer:

Fx  = 4.92 [N]

Fy = 0.868 [N]

Explanation:

Let's take the 10 degrees as a measure from the horizontal component to the vector.

Thus taking the components in the X & y axes respectively:

Fx = 5*cos(10) = 4.92 [N]

Fy = 5*sin(10) = 0.868 [N]

3 0
3 years ago
What is meant by pressure?write its unit.
Mars2501 [29]

Answer:

Pressure is the force applied perpendicular to the surface of an object per unit area over which that force is distributed

Unit: Pascal (Pa)

1Pa = 1N/m^2

5 0
3 years ago
A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th
NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2

v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)

v_{pf}=-31.4\ m/s

v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}

v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}

v_{gf} = 0.0988\ m/s

4 0
3 years ago
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