All categories

3 years ago

Antagonistic muscles move a body part in _____. 1. the same direction

Physics

2 answers:

alexdok [17]3 years ago

5 0

Answer: Opposite direction

Skeletal muscles arise in antagonistic pairs where the muscles contract to produce opposite movements at the same joint. Antagonistic muscles move a body part in one direction by contraction, the other moves that part in opposite direction.

In addition, when a muscle contracts to produced movement, its antagonist relaxes to allow movement to take place such as biceps muscles is a flexor muscle for elbow joint and triceps is the antagonist.

irga5000 [103]3 years ago

3 0

2. Opposite directions

You might be interested in

A baseball m=.34kg is spun vertically on a massless string of length l=.52m. the string can only support a tension of tmax=9.9n

larisa86 [58]

<span>4.5 m/s This is an exercise in centripetal force. The formula is F = mv^2/r where m = mass v = velocity r = radius Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop. Let's determine the force we get from gravity. 0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N Plug known values into formula. F = mv^2/r 13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m 6.88064 kg m^2/s^2 = 0.34 kg V^2 20.23717647 m^2/s^2 = V^2 4.498574938 m/s = V Rounding to 2 significant figures gives 4.5 m/s The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>

7 0

3 years ago

Read 2 more answers

In a machine shop, a hydraulic lift is used to raise heavy equipment for repairs. The system has a small piston with a cross-sec

Mrac [35]

Answer:

$F_s=1075.9493\ N$

Explanation:

Given:

area of piston on the smaller side of hydraulic lift, $a_s=0.075\ m^2$

area of piston on the larger side of hydraulic lift, $a_l=0.237\ m^2$

Weight of the engine on the larger side, $W_l=3400\ N$

Now, using Pascal's law which state that the pressure change in at any point in a confined continuum of an incompressible fluid is transmitted throughout the fluid at its each point.

$P_s=P_l$

$\frac{F_s}{a_s}=\frac{W_l}{a_l}$

$\frac{F_s}{0.075} =\frac{3400}{0.237}$

$F_s=1075.9493\ N$ is the required effort force.

5 0

3 years ago

Read 2 more answers

2. What is the difference between analytical response and concentration?

Anton [14]

An example helps clarify the difference between an analysis, a deter- mination and ... departments analyze samples of water to determine the concentration of ... moles of Cu2+, and cylinder 2 contains 20 mL, or 2.0 × 10.

5 0

3 years ago

A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20

statuscvo [17]

Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

Assuming no friction, total mechanical energy must keep constant, so the following is always true:

$\Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0 (1)$

Choosing the ground level as our zero reference level, Uf =0.
Since the child starts from rest, K₀ = 0.
From (1), ΔU becomes:
$\Delta U = 0- m*g*h = -m*g*h (2)$
In the same way, ΔK becomes:
$\Delta K = \frac{1}{2}*m*v_{1}^{2} (3)$
Replacing (2) and (3) in (1), and simplifying, we get:

$\frac{1}{2}*v_{1}^{2} = g*h (4)$

In order to find v₁, we need first to find h, the height of the slide.
From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

$h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)$

Replacing (5) in (4) and solving for v₁, we get:

$v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s (6)$

As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
We can do this in more than one way, but a very simple one is using kinematic equations.
If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

$v_{1}^{2} - v_{o}^{2} = 2*a* x_{1} (7)$

Since v₀ = 0 (the child starts from rest) we can solve for a:

$a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2 (8)$

Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

$t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s (9)$

Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

$x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2} (10)$

Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

$x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2} = 17.2 m (11)$

From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

$v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)$

Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

$\frac{1}{2}*v_{2}^{2} = g*h_{2} (13)$

Replacing from (12) in (13), we can solve for h₂:

$h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m (14)$

Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

$x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)$

4 0

2 years ago

Products and reactants contain the same _____ but in different combinations

Drupady [299]

Products and reactants contain the same atoms, but the atoms are in different combinations. Since the atoms are moved to other places in the reaction, a new chemical product is generally produced.

4 0

3 years ago