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creativ13 [48]
2 years ago
15

Answer plzzHow do we apply Interval Training to our lower body for a workout?

Physics
1 answer:
weqwewe [10]2 years ago
5 0
Interval training is simply alternating short bursts (about 30 seconds) of intense activity with longer intervals (about 1 to 2 minutes) of less intense activity. For instance, if your exercise is walking and you're in good shape, you might add short bursts of jogging into your regular brisk walks.
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A block of mass 5 kg is descending with a constant velocity on an inclined plane at 30°. What is the value of the frictional for
sattari [20]

Answer:28.9N

Explanation:

F=UR R=mg R=5×10 R=50N

U=tan© U=Tan30 U=1/√(3)

F=UR F=1/√(3) ×50

F=28.9N

4 0
2 years ago
An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t
Vlad1618 [11]

Answer:5

Explanation:

Given

speed of object v=1\ m/s

radius of circle r=1\ m

Force towards the center F=1\ N

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

W=F\cdot s\cos 90

W=0

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3 years ago
Naphthalene combustion can be used to calibrate the heat capacity of a bomb calorimeter. the heat of combustion of naphthalene i
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6 0
3 years ago
What variable is represented on the y-axis?
gladu [14]

Distance/ Time which means Distance is on horizontal and time is on vertical

8 0
3 years ago
A heavy piece of hanging sculpture is suspended by a 90 cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at it
kupik [55]

Answer: The mass of the sculpture is 11.8kg

Explanation:

Using the equation of fundamental frequency of a taut string.

f = (1/2L)*√(T/μ) .... (Eqn1)

Where

f= frequency in Hertz =80Hz

T = Tension in the string = Mg

M represent the mass of the substance (sculpture) =?

g= 9.8m/s^2

L= Length of the string=90cm=0.9m

μ= mass density = mass of string /Length of string

mass of string =5g=0.005kg

L=0.9m

μ=0.005/0.9 = 0.0056kg/m

Using (Eqn1)

80= 1/(2*0.9) √(T/0.0056)

144= √(T/0.0056)

Square both sides

20736= T/0.0056

T= 116.12N

Recall that T =Mg

116.12= M * 9.8

M=116.12/9.8

M= 11.8kg

Therefore the mass of the sculpture is 11.8kg

4 0
3 years ago
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