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Tpy6a [65]
3 years ago
10

Isla’s change in velocity is 30 m/s, and Hazel has the same change in velocity. Which best explains why they would have differen

t accelerations?
Physics
2 answers:
alexdok [17]3 years ago
6 0

Answer:

different time then hazel

Explanation:

lyudmila [28]3 years ago
4 0

Acceleration  =  (change in velocity) / (time for the change) .

They may have had the same change in velocity, but if the changes
happened in different lengths of time, then their accelerations were
not the same.
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Infer the direction of the net force acting on a car as it slows down and turns right.
Kobotan [32]
Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.
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Compare the foot of mollusks with the tube feet of echinoderms.
stira [4]

Mollusks live in fresh water, in marine environment, but also on land.

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3 years ago
Why are denser materials found closer to earths center
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Denser materials tend to be closer to earths center due to their mass gravity is shown by the equation mg
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3 years ago
The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.
irina [24]

Part a)

At t = 0  the position of the object is given as

x = 0

At t = 2

x = 2 sin(\pi/2) = 2cm

so displacement of the object is given as

d = 2 - 0 = 2cm

so average speed is given as

v_{avg} = \frac{2}{2} = 1 cm/s

Part b)

instantaneous speed is given by

v = \frac{dy}{dt}

v = 2cos(\pi t/4 ) * \frac{\pi}{4}

now at t= 0

v = \frac{\pi}{2} cm/s

at t = 1

v = 2 cos(\pi/4) * \frac{\pi}{4}

v = \frac{\pi}{2\sqrt2}

at t = 2

v = 0

Part c)

Average acceleration is given as

a_{avg} = \frac{v_f - v_i}{t}

a_{avg} = \frac{0 - \frac{\pi}{2}}{2}

a = -\frac{\pi}{4} cm/s^2

Part d)

Now for instantaneous acceleration

As we know that

a =- \omega^2 y

at t = 0

a = -\frac{\pi^2}{16} * 0 = 0 cm/s^2

at t = 1

y = \sqrt2 cm

now we have

a = -\frac{\pi^2}{16}*\sqrt2

At t = 2 we have

y = 2 cm

a = -\frac{\pi^2}{16}*2

a = -\frac{\pi^2}{8}

<em>so above is the instantaneous accelerations</em>

7 0
3 years ago
A 2.8 kg grinding wheel is in the form of a solid cylinder of radius 0.1 m. a) What constant torque will bring it from rest to a
kogti [31]

Answer:

a) τ =  0.672 N m , b) θ = 150 rad , c) W = 100.8 J

Explanation:

a) for this part let's start by finding angular acceleration, when the angular velocity stops it is zero (w = 0)

       w = w₀ + α t

       α = -w₀ / t

       α = 120 / 2.5

       α = 48 rad / s²

The moment of inertia of a cylinder is

       I = ½ M R²

Let's calculate the torque

      τ = I α

      τ = ½ M R² α

      τ = ½ 2.8 0.1² 48

      τ =  0.672 N m

b) we look for the angle by kinematics

      θ = w₀ t + ½ α t2

      θ = ½ α t²

      θ = ½ 48 2.5²

      θ = 150 rad

c) work in angular movement

      W = τ θ

      W = 0.672 150

      W = 100.8 J

7 0
3 years ago
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