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Strike441 [17]
3 years ago
9

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

a gazelle assumes an acceleration of 4.2 m/s2 for 6.5 s , after which the gazelle continues at a steady speed.
Physics
1 answer:
NikAS [45]3 years ago
6 0

Answer:

The gazelles top speed is 27.3 m/s.

Explanation:

Given that,

Acceleration = 4.2 m/s²

Time = 6.5 s

Suppose we need to find the gazelles top speed

The speed is equal to the product of acceleration and time.

We need to calculate the gazelles top speed

Using formula of speed

v=at

Where, v = speed

a = acceleration

t = time

Put the value into the formula

v=4.2\times6.5

v=27.3\ m/s

Hence, The gazelles top speed is 27.3 m/s.

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A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr
PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

where

\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 2.41\ Nm

 

3 0
3 years ago
Two identical bodies are sliding toward each other on a frictionless surface. One moves at 1 m/s and the other at 2 m/s. They co
mario62 [17]

Answer:

V=\dfrac{3}{2}\ m/s

Explanation:

Two identical bodies are sliding toward each other on a frictionless surface.

Initial speed of body 1, m₁ = 1 m/s

Initial speed of body 2, m₂ = 2 m/s

They collide and stick.

We need to find the speed of the combined mass. Let V is the speed of the combined mass.

Using the conservation of momentum.

m_1u_1+m_2u_2=V(m_1+m_2)

We have, m₁ = m₂ = m

m\times 1+m\times 2=(m+m)V\\\\m+2m=2m\times V\\\\3m=2mV\\\\V=\dfrac{3}{2}\ m/s

So, the speed of the combined mass is \dfrac{3}{2}\ m/s.

6 0
2 years ago
Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.
Alika [10]
Let k =  the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
     = 0.5*(k N/m)*(0.35 m)²
     = 0.06125k J

The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
     = 0.5*(1.8 kg)*(1.9 m/s)²
     = 3.249 J

Assume that negligible energy is lost when KE is converted into SE.
Therefore
0.06125k = 3.249
k = 53.04 N/m

Answer:  53 N/m  (nearest integer)

3 0
3 years ago
Why is the following situation impossible? A skater glides along a circular path. She defines a certain point on the circle as h
Arturiano [62]

Answer:

A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin.

So here in circular motion of the skater we can see that the total path length of the skater is along the arc of the circle while we can say that displacement is defined as the shortest distance between initial and final position of the object.

So it is not possible in any circle that arc-length is less than the chord joining the two points on the circle

As we know that arc length is given as

L = R\theta

length of chord is given as

L_c = 2Rsin(\frac{\theta}{2})

so here

L > L_c

R\theta > 2R sin(\frac{\theta}{2})

so we have

\frac{\theta}{2} > sin(\frac{\theta}{2})

6 0
3 years ago
Two forces act on a moving object that has a mass of 27 kg. One has a magnitude of 12 N and points due south, while the other ha
levacccp [35]

0.77 m/s2 directed 35° south of west

net force = (-17,-12)

net force = mass * acceleration

(-17,-12) = 27 * (x-acceleration,y-acceleration)

(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)

angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.

magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)

7 0
2 years ago
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