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3 years ago

A student carries a backpack for one mile. Another student carries the same backpack for two miles

Physics

1 answer:

Svetllana [295]3 years ago

3 0

Answer:

Compared to the first student, the second student did twice as much work as the first student.

Explanation:

The work done by the first student will be equal to the Force exerted by the backpack on the student carrying it multiplied by one mile (Distance). The work done by the second student will be equal to the Force exerted by the backpack on the student carrying it multiplied by two miles (Distance).

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A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that

lina2011 [118]

Answer:

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

5 0

2 years ago

NGSS Physics

vovangra [49]

Answer: 3.5 seconds

EXPLANATION:

Using the formula:
v = u + at
And taking the upwards direction as positive, we have the following information:

u = 35 m/s
a = -10m/s^2 (this is acceleration due to gravity)

At the top of its path, the apple will have a velocity of 0 m/s, therefore:

v = 0m/s

Once you substitute everything into the formula, you get:

0 = 35 + (-10)t

Therefore, t = 35/10 = 3.5 seconds

6 0

3 years ago

Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius

Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

$a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2$

The centripetal acceleration for the second radius; 4.0 m

$a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2$

The centripetal acceleration for the third radius; 6.0 m

$a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2$

The centripetal acceleration for the fourth radius; 8.0 m

$a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2$

The centripetal acceleration for the fifth radius; 10.0 m

$a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2$

6 0

3 years ago

The High Speed Industrial Drill With Diameter Of 98 Cm Develops 5.85hp At 1900 Rpm. What Torque And Force Is Applied To The Dril

STatiana [176]

Answer:

The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

Explanation:

Given:-

- The diameter of the drill bit, d = 98 cm

- The power at which drill works, P = 5.85 hp

- The rotational speed of drill, N = 1900 rpm

Find:-

What Torque And Force Is Applied To The Drill Bit?

Solution:-

- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).

- The relation between these quantities is given:

T = 5252*P / N

T = 5252*5.85 / 1900

T = 16.171 Nm

- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):

T = F*r

Where, r = d / 2

F = 2T / d

F = 2*16.171 / 0.98

F = 33 N

Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.

4 0

3 years ago

Read 2 more answers

A grandfather clock depends on the period of a pendulum to keep correct time.(ii) Suppose a grandfather clock is calibrated corr

ANEK [815]

The grandfather clock will now run slow (Option A).

<h3>What is Time Period of an oscillation?</h3>

The time period of an oscillation refers to the time taken by an object to complete one oscillation.
It is the inverse of frequency of oscillation; denoted by "T".

Now,

$\sqrt{\frac{L}{g}}$ , where L is the length and g is the gravitational constant, is the formula for a pendulum's period.
The period will increase as one climbs a very tall mountain because g will slightly decrease.
Due to this and the previous issue, the clock runs slowly and it seems that one second is longer than it actually is.

Hence, the grandfather clock will now run slow (Option A).

To learn more about the time period of an oscillation, refer to the link: brainly.com/question/26449711

#SPJ4

6 0

1 year ago