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goldfiish [28.3K]
4 years ago
9

Using the rules for significant figures what do you get when you multiply 67.6 by 1.2

Physics
2 answers:
stira [4]4 years ago
6 0

When you multiply a significant figure the answer should never be more digits then the number with the least amount of digits so 67.6 is 3 digits and 1.2 is 2 digits so the answer cannot be more then 2 digits in this case

The Answer is 81

Deffense [45]4 years ago
4 0
811.2 , hope this helps and please mark me as brainlyest
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Every summer I drive to Pennsylvania. It is 895 km to get there. If I average 100 km/hr, how much time will I spend driving?
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Answer the time you bee spending driving iss 795 because 895-795=100

Explanation:

6 0
2 years ago
If Vector A is (3, 0) and Vector B is (-3, 3), what is the resultant?
Archy [21]

Answer:resultant vector R = (0, 3)

Explanation: vector A = (3, 0)

vector B =(-3, 3)

Vectors are added such that those in same directions are added together. The resultant vector R is the given by R = (3-3, 0+3)

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3 years ago
A 3.0 kg box is suspended by a series of ropes as shown below. The tension force in the horizontal rope is -40 N. What is the te
Nitella [24]

Answer:

50 N

Explanation:

Let the force in the horizontal rope be F₁ and the force in the diagonal rope be F₂:

The total force in the horizontal and vertical directions must be zero, since the object is at rest and is not accelerating.

The horizontal component of the forces:

F₁ + F₂ = -40N + F₂ = 0

F₂ = 40N

The vertical component of the forces:

F₁ + F₂ - mg = 0 + F₂ - mg = 0

F₂ = mg

If I assume the gravitational constant g = 10 m/s²:

F₂ = (3 kg) * (10 m/s²) = 30N

Adding the horizontal and vertical components of the force F₂:

F₂ = √((40N)² + (30N)²) = 50N

6 0
3 years ago
An electric motor consumes 9.00 kj of electrical energy in 1.00 min. if one-third of this energy goes into heat and other forms
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5 0
3 years ago
A 50.0 kg crate is pulled 375 N of force applied to a rope. The crate slides without friction.
LUCKY_DIMON [66]

Hi there!

We can use the work-energy theorem to solve.

Recall that:

\large\boxed{W = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2}

The initial kinetic energy is 0 J because the crate begins from rest, so we can plug in the given values for mass and final velocity:

W = \frac{1}{2}(50)(5.61^2) = 786.8025 J

Now, we can define work:

\large\boxed{W = Fdcos\theta}}

Now, plug in the values:

786.8025 = Fdcos\theta\\\\786.8025 = (375)(3.07)cos\theta

Solve for theta:

cos\theta = .6834\\\theta = cos^{-1}(.6834) = \boxed{46.887^o}

4 0
3 years ago
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