PHYSICS MIDDLE SCHOOL

Adam pushes a 47 kg boat across a frozen lake. The coefficient of kinetic friction between the boat and the ice is 0.13 and the boat acceleration at 0.6 m/s^2. Calculate the magnitude of Adams force in the boat.

Answers

Answer 1

Answer:

Answer:

The magnitude of Adams' force is 88 N

Explanation:

According to Newton's law ∑ forces in direction of motion is equal to

mass multiplied by the acceleration

There are two forces here the friction force and Adams' force

Adams' force is in direction of motion

Friction force is opposite to direction of motion

→ Friction force = μ R,

where μ is the coefficient of friction and R is the normal reaction force

of the boat

→ R = mg

where m is the mass of the boat and g is the acceleration of gravity

→ m = 47 kg , g = 9.8 m/s²

→ R = mg

Substitute the values in the rule

→ R = 47 × 9.8 = 460.6 N

→ Friction force = μ R

→ μ = 0.13 , R = 460.0 N

Substitute the values in the rule

→ Friction force = 0.13 × 460.0 = 59.878 N

→ ∑ Forces in direction of motion = mass × acceleration

→ F - Friction force = mass × acceleration

→ Friction force = 59.878 N , m = 47 kg , a = 0.6 m/s²

Substitute the values in the rule

→ F - 59.878 = 47 × 0.6

→ F - 59.878 = 28.2

Add 59.878 to both sides

→ F = 88.078 N ≅ 88 N

The magnitude of Adams' force is 88 N

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dayshawn is traveling at 12 m/s away from the school. dayshaun's mom is looking for him because he is late coming home so she left the house and is driving toward the school at 5 m/s. The school is 6492 m from dayshauns house. at what position will dayshawn meet his mom and how long after she left the house did she find dayshaun? draw a motion map and solve. show your work

Answers

Speed of Dayshawn travelling towards his home is 12 m/s

Speed of her mom towards his school is 5 m/s

They both starts at same time so whenever they will meet on their path the sum of the distance covered by Dayshawn and distance covered by his mom must be equal to the total distance of school and home

Now let say they both meet after "t" time when they starts motion

so we can write the total distance between school and home as

$d = v_1*t + v_2*t$

here d = 6492 m

$v_1 = 12 m/s$ = speed of dayshawn

$v_2 = 5 m/s$ = speed of his mom

now by solving the above equation

$6492 = 12t + 5 t$

$t = (6492)/(17)$

$t = 381.9 s$

so they will meet after 381.9 s from start which will be 3.36 minutes from there start

Also at this time the distance covered by her mom will be

$d_2 = v_2*t$

$d_2 = 5* 381.9 = 1909.4 m$

so they will meet at distance 1909.4 m from their home

The angle of incidence and the angle of reflection are measured from the_____, the perpendicular to the surface.

Answers

I believe it is angles

3. A block of mass m1=1.5 kg on an inclined plane of an angle of 12° is connected by a cord over a mass-less, frictionless pulley to a second block of a mass m2= 1.2 kg hanging vertically. If the hanging block falls 0.92min 1.23 s, what is the coefficient of friction between m1 and the inclined plane?

Please show work

Answers

Answer:

$\mu=0.377$

Explanation:

we need to start by drawing the free body diagram for each of the masses in the system. Please see attached image for reference.

We have identified in green the forces on the blocks due to acceleration of gravity ( $w_1$ and $w_2$ ) which equal the product of the block's mass times "g".

On the second block ( $m_2$ ), there are just two forces acting: the block's weight ( $m_2\,*\,g$ ) and the tension (T) of the string. We know that this block is being accelerated since it has fallen 0.92 m in 1.23 seconds. We can find its acceleration with this information, and then use it to find the value of the string's tension (T). We would need both these values to set the systems of equations for block 1 in order to find the requested coefficient of friction.

To find the acceleration of block 2 (which by the way is the same acceleration that block 1 has since the string doesn't stretch) we use kinematics of an accelerated object, making use of the info on distance it fell (0.92 m) in the given time (1.23 s):

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2$ and assume there was no initial velocity imparted to the block:

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2\n-0.92\,m=0\,-(1)/(2) a\,(1.23)^2\na=(0.92\,*\,2)/(1.23^2) \na=1.216 \,(m)/(s^2)$

Now we use Newton's second law in block 2, stating that the net force in the block equals the block's mass times its acceleration:

$F_(net)=m_2\,a\nw_2-T=m_2\,a\nm_2\,g-T=m_2\,a\nm_2\,g-m_2\,a=T\nm_2\,(g-a)=T\n1.2\,(9.8-1.216)\,N=T\nT=10.3008\,N$

We can round this tension (T) value to 10.3 N to make our calculations easier.

Now, with the info obtained with block 2 (a - 1.216 $(m)/(s^2)$ , and T = 10.3 N), we can set Newton's second law equations for block 1.

To make our study easier, we study forces in a coordinate system with the x-axis parallel to the inclined plane, and the y-axis perpendicular to it. This way, the motion in the y axis is driven by the y-component of mass' 1 weight (weight1 times cos(12) -represented with a thin grey trace in the image) and the normal force (n picture in blue in the image) exerted by the plane on the block. We know there is no acceleration or movement of the block in this direction (the normal and the x-component of the weight cancel each other out), so we can determine the value of the normal force (n):

$n-m_1\,g\,cos(12^o)=0\nn=m_1\,g\,cos(12^o)\nn=1.5\,*\,9.8\,cos(12^o)\nn=14.38\,N$

Now we can set the more complex Newton's second law for the net force acting on the x-axis for this block. Pointing towards the pulley (direction of the resultant acceleration $a$ ), we have the string's tension (T). Pointing in the opposite direction we have two forces: the force of friction (f ) with the plane, and the x-axis component of the block's weight (weight1 times sin(12)):

$F_(net)=m_1\,a\nT-f-w_1\,sin(12)=m_1\,a\nT-w_1\,sin(12)-m_1\,a=f\nf=[10.3-1.5\,*\,9.8\,sin(12)-1.5\,*1.216]\,N\nf=5.42\,N$

And now, we recall that the force of friction equals the product of the coefficient of friction (our unknown $\mu$ ) times the magnitude of the normal force (14.38 N):

$f=\mu\,n\n5.42\,N=\mu\,*\,14.38\,N\n\mu=(5.42)/(14.38)\n\mu=0.377$

with no units.

1)the magnetic field is strongest near_________of a bar magnet.a)the poles b)the center

2)the needle of compass line up with earth magnetic field and point to_________
a)earth poles b)earth equator

3)magnetic field lines that curve toward each other show_________
a)repulsion b)attraction

4)some animals have tiny pieces of ___________in their brains to help them find their way
a)magnetite b)magnetosphere

5)a magnet contain a large number of magnetic _____that are lined up and pointed in the same direction.
a)domains b)poles

Answers

Answer:

1 . a) the poles

2. a) earth poles.

3. b) attraction.

4. a) magnetite.

5. a) domains.

Explanation:

The magnetic field is strongest at the poles and very weak at the center of the magnet.

The magnetic compass will always align along the north - south poles of the earth. North of the compass needle is attracted towards the magnetic south pole of the earth.

Magnetic Field lines begin at the north pole of the magnet and end at the south pole. Unlike poles attract and like poles repel , so they curve for unlike poles.

Magnetite is a magnetic mineral, which is present in the animal brains , enabling them to identify correct directions.

Magnetic domains are regions where individual magnetic moments of the atoms are aligned in a definite direction.

1. Is A. at the poles because thats where the magnetic field is going out then coming back into the earth to produce the magnetic field.
2. Again its A. because the compass needle is attracted to " north " which is magnetic south. It does this because opposites attract.
3. This one would be B. Because if the magnets were being repelled the magnetic field lines would look like there was a line that the field hit and bounced off of it.
4. This answer is A. the magnetite helps them migrate so they know which way is north and which way is south.
5. This answer is A. Because without the domains there wouldn't be poles on the magnetic object.

What structure help support plants

Answers

The cell structure that helps support plants is the cell wall. It provides a shell structure around the main cell of the plant and its layers so then it will have the adequate amount of strength that it needs. It's also one of the reasons why even non-woody plants such as herbs have the strength of support itself.

The structure that help support plants is the stem

What is a planet that formed closer to the Sun than Earth but not closest to the sun

Answers

Venus and Mars are both closer to the sun than earth however, neither one is the closest. The planet Mercury is closest to the sun.Fun fact! Did you know that The brightest looking object in the the sky at night is Venus? It appears to be an overly bright star due to the reflection of the sun.For brainiest answer, choose what you think is the best answer!

Venus or Mars. Please nominate me Brainiest when someone else answers for my quick responding and accuracy. Thanks and have a good school year!

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