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3 years ago

Starting from rest and moving in a straight line, a cheetah

Physics

1 answer:

UNO [17]3 years ago

3 0

The average acceleration of the cheetah is 7.75 m/s²

The given parameters;

velocity of the cheetah, v = 31 m/s

time of motion, t = 4 seconds

The average acceleration of the cheetah is calculated as follows;

$Average \ acceleration = \frac{\Delta velocity}{\Delta time } \\\\Average \ acceleration = \frac{v-u}{t}$

where;

v is the final velocity

u is the initial velocity

t is the time of motion

Assuming the cheetah started from rest, the initial velocity, u = 0

The average acceleration is calculated as follows;

$Average \ acceleration = \frac{v-u}{t} = \frac{31-0}{4} = 7.75 \ m/s^2$

Thus, the average acceleration of the cheetah is 7.75 m/s²

Learn more here: brainly.com/question/17280180

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In 1896, S. Riva-Rocci developed the prototype of the current sphygmomanometer, a device used to measure blood pressure. When it

Anastasy [175]

Answer:

Explanation:

A pressure that causes the Hg column to rise 1 millimeter is called a torr. The term 1 mmHg used can replaced by the torr.

1 atm = 760 torr = 14.7 psi.

120 mmHg

Psi:

760 mmHg = 14.7 psi

120 mmHg = 14.7/760 * 120

= 2.32 psi

Pa:

1mmHg = 133.322 Pa

120 mmHg = 120 * 133.322

= 15998.4 Pa

80 mmHg

Psi:

760 mmHg = 14.7 psi

80 mmHg = 14.7/760 * 80

= 1.55 psi

Pa:

1mmHg = 133.322 Pa

80 mmHg = 80 * 133.322

= 10665.6 Pa

4 0

3 years ago

In a meeting of mimes, mime 1 goes through a displacement d1 = (6.00 m)i + (5.74 m)j and and mime 2 goes through a displacement

Blizzard [7]

a) Vector product: |d1 × d2| = (34.2 m) k

b) Scalar product: d1 · d2 = -0.874 m

c) (d1 + d2) · d2 = 16.1 m

d) Component of d1 along direction of d2: -0.21 m

Explanation:

In this part, we want to calculate

|d1 × d2|

Which is the vector product between the two displacements d1 and d2.

The two vectors are:

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

The vector product of two vectors $(a_x,a_y,a_z)$ and $(b_x,b_y,b_z)$ is also a vector which has components:

$r=(r_x,r_y,r_z)\\r_x=a_yb_z-a_zb_y\\r_y=a_zb_x-a_xb_z\\r_z=a_xb_y-a_yb_x$

We notice immediatly that in this problem ,the two vectors d1 and d2 lie in the x-y plane, so they do not have components in zero. Therefore, the vector product has only one component, which is the one in z, and it is:

$r_z=(6.00)(2.9)-(5.74)(-2.92))=34.2 m$

Therefore, the vector product of d1 and d2 is:

|d1 × d2| = (34.2 m) k

In this case, we want to calculate

d1 · d2

Which is the scalar product between the two displacements.

The scalar product of two vectors $(a_x,a_y,a_z)$ and $(b_x,b_y,b_z)$ is a scalar given by:

$a \cdot b = a_x b_x + a_y b_y + a_z b_z$

In this problem,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

Therefore, the scalar product between the two vectors is:

$d_1 \cdot d_2 = (6.00)(-2.92)+(5.74)(2.9)=-0.87m$

In this case, we want to calculate

(d1 + d2) · d2

Which means that first we have to calculate the resultant displacement d1 + d2, and then calculate the scalar product of the resultant vector with d2.

Given two vectors $(a_x,a_y,a_z)$ and $(b_x,b_y,b_z)$ , the resultant vector is also a vector given by

$r=(r_x,r_y,r_z)\\r_x=a_x+b_x\\r_y=a_y+b_y\\r_z=a_z+b_z$

In this case,

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

So the resultant vector is

$r_x=6.00+(-2.92)=3.08 m\\r_y = 5.74+2.9=8.64 m$

$(d_1+d_2)=(3.08 m,8.64 m)$

And calculating the scalar product with d2, we find:

$(d_1 + d_2)\cdot d_2 = (3.08)(-2.92)+(8.64)(2.9)=16.1 m$

The component of a vector a along another vector b is given by

$a_b = \frac{a\cdot b}{|b|}$

where $a\cdot b$ is the scalar product between and b

$|b|$ is the magnitude of vector b

In this problem, we have the two vectors

d1 = (6.00 m)i + (5.74 m)j

d2 = (-2.92 m)i + (2.9 m)j

We want to find the component of d1 along the direction of d2.

We already calculated the scalar product of the two vectors in part b):

d1 · d2 = -0.874 m

The magnitude of a vector b is given by

$|b|=\sqrt{b_x^2+b_y^2+b_z^2}$

So, for vector d2,

$|d_2|=\sqrt{(-2.92)^2+(2.9)^2}=4.1 m$

Now we can calculate the component of d1 along d2:

$d_1_{d_2}=\frac{d_1 \cdot d_2}{|d_2|}=\frac{-0.874}{4.1}=-0.21 m$

Learn more about operations with vectors:

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4 0

4 years ago

Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a

lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

5 0

4 years ago

How is velocity ratio of wheel and axle calculated

Rudiy27

Answer:

VR = $\frac{Radius of the wheel}{Radius of the axle}$

Explanation:

Velocity ratio (VR) of a machine is a term that compares the distance moved by effort put into the machine to the distance moved by the load.

A wheel and axle is a device for lifting of a load through a height. It is made up of two circular parts called wheel and axle. Its velocity ratio (VR) can be determined by:

VR = $\frac{Radius of the wheel}{Radius of the axle}$

For a practical wheel and axle, the diameter of the wheel is greater than the diameter of the axle.

7 0

3 years ago

Is the answer D?

irga5000 [103]

Nope answer is 500 N downward
draw the freebody diagram to understand

7 0

4 years ago