4. There are two tree frogs that have different proteins in their cells for tongue length. They have

During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive

olganol [36]

Answer:

The force is $F_c = 789.03 \ N$

Explanation:

From the question we are told that

The tangential resistive force is $F_t = 115 \ N$

The mass of the wheel is m = 1.80 kg

The diameter of the wheel is $d = 50.0 cm = 0.5 \ m$

The diameter of the sprocket is $d_c = 8.50 \ cm =0.085 \ m$

The angular acceleration considered is $\alpha = 4.30\ rad/s^2$

Generally the radius of the wheel is

$r = \frac{d}{2}$

=> $r = \frac{0.5}{2}$

=> $r = 0.25 \ m$

Generally the radius of the sprocket is

$r_c = \frac{d_c}{2}$

=> $r_c = \frac{0.085}{2}$

=> $r_c = 0.0425 \ m$

Generally the moment of inertia of the wheel is mathematically represented as

$I = m * r^2$

=> $I = 1.80 * 0.25^2$

=> $I = 1.1125 \ kg \cdot m^2$

Generally the torque experienced by the wheel due to the forces acting on it is mathematically represented as

$\tau = F_c * r_c - F_t * r$

Here $F_c$ is the force acting on the sprocket

So

$\tau = F_c * 0.0425 - 115 * 0.25$

$\tau = 0.0425F_c - 28.75$

Generally the torques that will cause the wheel to move with $\alpha = 4.30\ rad/s^2$ is mathematically represented as

$\tau = I * \alpha$

So

$0.0425F_c - 28.75 = I * \alpha$

$0.0425F_c - 28.75 = 1.1125 *4.30$

$F_c = 789.03 \ N$

5 0

3 years ago

10. Bites of some animals can cause sprain tetanus rables

Burka [1]

Answer:

an infection, including bacterial and rabies infections.

broken animal teeth embedded in your wound.

foreign objects embedded in your wound.

possible nerve and blood vessel damage.

5 0

3 years ago

A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. What is the maximum height h that it achieves, and w

sveta [45]

Answer:

A) t = 0.40816 s , y = 0.916 m

Explanation:

A) For this problem we use the kinematic relations

v = v₀ - g t

the highest point zero velocities (v = 0)

t = (v₀-v) / g

t = (4 - 0) / 9.8

t = 0.40816 s

to calculate the height let's use

v² = v₀² - 2 g y

y = vo2 / 2g

y = 4 2 / (2 9.8)

y = 0.916 m

To find the number of photos, we can use a direct proportions rule, if you take 30 photos in a second in 0.40816 s how many photos does it take

# _photos1 = 0.40916 (30/1)

# _photos1 = 12

yes i take 120 fps

#_fotod = 0.40916 (120/1)

#photos = 5.87 10³

B) The ball is released from a latura h how long it takes to reach the floor

v² = v₀² + 2 g y

where the initial velocity is zero and the velocity with which the expert leaves is equal to the velocity with which v = 4 m / s leaves

v² = 2gy

v = √ (2 9.8 0.916)

v = √ (2.1397 101)

v = 4.6257 m / s

c) we ask us for the time for latura

y = L / 2

y = 0.916 / 2

y = 0.458 m

now we can use the formula

y = v₀ t - ½ g t²

0.458 = 4.00 t - ½ 9.8 t²

4.9 t² - 4t + 0.458 = 0

t² -0.8163 t +0.09346 = 0

we solve second degree execution

t = [0.8163 ±√ (0.8163² - 4 0.09346)] / 2

t = [0.8163 ± 0.540] / 2

t₁ = 0.678 m

t₂ = 0.2763 m

the shortest time is for when the ball goes up and the longest when it goes down

D) the graph of vs Vs is expected to be a closed line

and the graph of position versus time a parabola

0 0

3 years ago

For a vehicle to negotiate a banked curve in poor weather conditions, where the force of friction f = 0, for a given velocity v

djverab [1.8K]

Answer:

$R=240m$

Explanation:

From the question we are told that:

Velocity $v=25m/s$

Force of friction f = 0

Angle $\theta=15$

Generally the equation for Radius of curvature is mathematically given by

$R=frac{v^2}{tan\theta *g}$

$R=frac{25^2}{tan 15 *9.81}$

$R=240m$

7 0

3 years ago

A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is

Valentin [98]

Answer:

The end of the ramp is 38.416 m high

Explanation:

Horizontal Motion

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

If the maximum horizontal distance is known, we can solve the above equation for h:

$\displaystyle h=\frac {d^2g}{2v^2}$

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

$\displaystyle h=\frac {70^2\cdot 9.8}{2\cdot 25^2}$

$\displaystyle h=\frac {4900\cdot 9.8}{2\cdot 625}$

h= 38.416 m

The end of the ramp is 38.416 m high

8 0

3 years ago