All categories

3 years ago

A horse ran at a constant speed for 4 hours. Then, it decreased its speed by 7 mph for the

Physics

1 answer:

algol133 years ago

6 0

Distance travelled in 4 hours...

730 - 442 = 288 miles

Speed for the 4 hours;

Speed = 288/4 = 72 miles/hour

Speed for the rest of 442 miles...;

(72 - 7) = 65 miles/hour

Time taken for 442 miles..

Time taken = Distance/Speed

; 442/65 = 6.8

Time taken = 6.8 hours

You might be interested in

Kendall has empty graduated cylinder with markings and an identical graduated cylinder partway filled with water. She also has a

Elis [28]

The answer is B.
She can measure the mass of the water, marble and the graduated cylinder with the balance.
The volume of the water can be shown on the marked graduated cylinder, the volume of the marble can be measured by the volume difference of the water before and after the marble is put in.

4 0

2 years ago

Read 2 more answers

) The square plates of a 5000-pF parallel-plate capacitor measure 50 mm by 50 mm and are separated by a dielectric that is 0.23

notsponge [240]

Answer:

4 x 10⁻⁴ J

Explanation:

C = 5000 pF, V = 400 V

Energy = CV²/2 = 5000 x 10⁻¹² x 400²/2 = 4 x 10⁻⁴ J

6 0

2 years ago

In operant conditioning, many complex behaviors are learned through shaping. T or F

Snezhnost [94]

Answer:

true

Explanation:

8 0

3 years ago

Is the earth clean...? <br> A. hell no! <br> B. YES<br> C. I dont know.

Tamiku [17]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Dete

jeka57 [31]

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

$y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m$

b) the mass moment of inertia about z axis passing the rotation center O is:

$I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2$

c) The mass moment of inertia about z axis passing the rotation center O is:

$I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2$

3 0

3 years ago