Ask question

Ask question

All categories

3 years ago

8

John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho

vel. He is pushing down on the opposite end of the shovel with one hand and holding it up 30.0 cm from that end with his other hand. Ignore the mass of the shovel. Sarah says that his hand pushing down on the shovel must be exerting a greater force than the hand pushing up. James says it is just the reverse. Which one, if either, is correct?

1 answer:

Andru [333]3 years ago

8 0

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

$F_{down} L_1 = W_{snow} L_2$

so here we have

$F_{down} = \frac{L_2}{L_1} (W_{snow})$

so here we can say that upward force by which we push up is always more than the downwards force

You might be interested in

Ill mark brainliest!!!

navik [9.2K]

Answer: A

si, di, are expressed as negative and f is expressed as positive values.

Explanation:

When the object is located at a location in front of the focal point of a convex len, the image will always be located somewhere on the same side of the lens as the object. The image is located behind the object. In this case, the image will be an upright image and the image is enlarged

In this question, the relationship between the focal length, image distance di and object distance is

I/f = 1/di + 1/do that is

I/f - 1/di - 1/do = 0

The image size si is also negative since the image is a virtual image.

Therefore, si, di, are expressed as negative, f is expressed as positive values.

6 0

3 years ago

The amount of work done depends on what two things

fomenos

mass and acceleration have a nice day

8 0

3 years ago

Help please I’ll mark as brainliest

Igoryamba

Answer:

just find the answer in Google because I post it into Google

7 0

3 years ago

Using Kepler's 3rd law and Newton's law of universal gravitation, find the period of revolution P of the planet as it moves arou

weeeeeb [17]

Answer:

$P = \pi \sqrt{\frac{(R_1+R_2)^3}{2 \ GMS}}$

Explanation:

From the attached diagram below:

AC = a (1 + e) = R₂ -------- equation (1)

CD = a ( 1 - e) = R₁ --------- equation (2)

⇒ 1 - e = $\frac{R_1}{a}$

$e= 1 - \frac{R_1}{a}$

Replacing the value for e into equation (1)

$a(1+1- \frac{R_1}{a})= R_2$

$= 2a - R_1 = R_2$

$a= \frac{R_1+R_2}{2}$

From Kepler's third law;

$P = 2 \pi \sqrt{\frac{a^3}{GMS}}$

$P = 2 \pi \sqrt{\frac{(R_1+R_2)^3}{8GMS}}$

$P = \pi \sqrt{\frac{(R_1+R_2)^3}{2 \ GMS}}$

5 0

3 years ago

A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. W

mina [271]

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

Solution:

As per the question:

Mass of the boulder, $m_{B} = 1000\ kg$

Mass of the sled, $m_{S} = 2.50\ kg$

Mass of the boy, $m_{b} = 40\ kg$

Initial Velocity, v = 10.0 m/s

Now,

To calculate the speed of both the sled and the boulder after the occurrence of the collision:

m = $m_{b} + m_{S} = 40 + 2.50 = 42.50\ kg$

Initial velocity of the boulder, $v_{B} = 0\ m/s$

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

$mv + m_{B}v_{B} = mv' + m_{B}v'_{B}$

where

v' = final velocity of the the system of boy and sled

$v'_{B}$ = final velocity of the boulder

$42.50\times 10 + m_{B}.0 = 42.50v' + 1000v'_{B}$

$42.50v' + 1000v'_{B} = 425$ (1)

Now,

Using conservation of energy:

$\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}$

$42.50\times 10^{2} + m_{B}.0 = 42.50v'^{2} + 1000v'_{B}^{2}$

$42.50v'^{2} + 1000v'_{B}^{2} = 4250$ (2)

Now, from eqn (1) and (2):

$v' = \frac{m - m_{B}}{m + m_{B}}\times v$

$v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s$

Now,

$v'_{B} = \frac{2m}{m + m_{B}}\times v$

$v'_{B} = \frac{2\times 42.50}{42.5 + 1000}\times 10 = 0.82\ m/s$

5 0

3 years ago

Read 2 more answers