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adoni [48]
3 years ago
12

A boat moves with a speed of +2.5 m/s in a direction 25° north of east. If the mass of the boat is 15,000 kg, what is the moment

um in the northward direction?
Physics
1 answer:
nadya68 [22]3 years ago
5 0
Momentum - mass in motion
P=MV
P=(15,000 kg)(2.5 m/s)
P=37 500 kg x m/s to the north
Hope this helps
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A motorcycle accelerates uniformly from rest at 7.9\,\dfrac{\text{m}}{\text{s}^2}7.9 s 2 m ​ 7, point, 9, space, start fraction,
8090 [49]

Answer:

t = 3.516 s

Explanation:

The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:

v(t) = v_0 +at

Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0

If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.

This can be done knowing that

1 km = 1000 m

1 h = 3600 s

Therefore

1 km/h = (1000/3600) m/s = 0.2777... m/s

100 km/h = 27.777... m/s

Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:

27.777 m/s = 7.9 m/s^2 t

Solving for t

t = (27.7777 / 7.9) s = 3.516 s

6 0
3 years ago
The emt must assume that any unwitnessed water-related incident is accompanied by:________
natta225 [31]

The EMT must assume that any unwitnessed water-related incident is accompanied by potential spinal damage.

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3 0
2 years ago
What is kinetic energy?
Lemur [1.5K]

Answer:

A, the energy an object has due to its motion.

Explanation:

Kinetic energy is the energy created by motion.

3 0
3 years ago
Read 2 more answers
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a
Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration=88.6m/s^2

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution=2\pi rad

2.59 rev=2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s

By using \pi=3.14

Angular velocity=\omega=16.27rad/s

Distance from axis=r=0.329 m

Tangential speed=r\omega=16.27\times 0.329=5.4m/s

Radial acceleration=\frac{v^2}{r}

Radial acceleration=\frac{(5.4)^2}{0.329}=88.6m/s^2

6 0
3 years ago
A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k
adell [148]

Answer:

a) = 10.22 rad/s

b) = 0.35 m

Explanation:

Given

Mass of the particle, m = 1.1 kg

Force constant of the spring, k = 115 N/m

Distance at which the mass is released, d = 0.35 m

According to the differential equation of s Simple Harmonic Motion,

ω² = k / m, where

ω = angular frequency in rad/s

k = force constant in N/m

m = mass in kg

So,

ω² = 115 / 1.1

ω² = 104.55

ω = √104.55

ω = 10.22 rad/s

If y(0) = -0.35 m and we want our A to be positive, then suffice to say,

The value of coefficient A in meters is 0.35 m

6 0
3 years ago
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