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Sergeu [11.5K]
2 years ago
11

You walk 12.0 m West and then 4.00 m south. What is your displacement?​

Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
7 0

Answer:

12.6 (3 sig. fig.)

Explanation:

(12^2 + 4^2)^1/2 = 12.6 (3 sig. fig.)

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Please indicate how long each bar is in centimeters and millimeters.
tatuchka [14]

Answer:

1). 9 cm

90 mm

2). 94 cm

940 mm

3). 27 cm

270 mm

4). 87 cm

870 mm

8 0
2 years ago
What is the maximum speed (in units of m/s) with which a car can round a
IrinaK [193]

Answer:

B)  15.5 m/s

Explanation:

r = 60m

μs = 0.4

using the formula max V = √r*g*μs (flat roadway)

v = sqrt(60 * 10 * 0.4)

v = 15.5 m/s

5 0
3 years ago
You are performing an experiment in which you measure the difference in
goblinko [34]
A. Thermometer would be the answer because you’re measuring temperature
3 0
2 years ago
Can there be any heat transfer between two bodies that are at the same temperature but at different pressures?
Alenkinab [10]

Answer:

No

Explanation:

When the two bodies at different temperatures then heat transfer takes place between two bodies. Temperature is the necessary condition for the heat transfer.

The energy can flow due to pressure difference but the heat transfer can not take place only due to pressure difference. Heat transfer needs temperature difference.

Therefore the answer will be No.

3 0
3 years ago
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
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