A boy is pulling his two sisters on a sled.
1 answer:
Answer:
Explanation:
The oly way we can figure this out is if the boy is pulling the sled at a constant velocity. If not, we need a value for acceleration, and you don't have that here. If the boy is pulling the sled at a constant velocity, then the value for acceleration is 0, making this a really simple problem. I'm going with that, since there is no way to answer you otherwise. If velocity is not constant, please either repost the question or put it in the notes section under the question as it stands. If acceleration is 0, then
F - f = ma becomes
F - f = m(0) which is
F - f = 0 and
F = f which says that the applied force is the same as the frictional force. We need then to find the frictional force, which has an equation of
f = μ where normal force is the same as the weight of the 2 girls. We will find that, then:
Each girl's mass is different so the normal force/weight equation is
w = (30.0)(9.8) + (40.0)(9.8) to get
w = 290 + 390 and
w = 680. Plug that into the frictional force equation:
f = (.120)(680) so
f = 82N
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Answer:
just before landing the ground
Explanation:
Let the velocity of projection is u and the angle of projection is 30°.
Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.
initial horizontal component of velocity, ux = u Cos 30
initial vertical component of velocity, uy = u Sin 30
Time of flight is given by
Final horizontal component of velocity, vx = ux = u Cos 30
Let vy is teh final vertical component of velocity.
Use first equation of motion
vy = uy - gT
vy = - u Sin 30
The magnitude of final velocity is given by
v = u
Thus, the velocity is same as it just reaches the ground.
Answer:
The magnitude of Force is 8.58×10⁵N and direction is upwards
Explanation:
The work beam does on the pile driver is given by
W=(FCos180°)Δx= -F(0.088m)
From work energy theorem
Choosing y=0 at the the level where the driver first contacts the beam and vi=0 at yi=+3.40m and comes to rest again vf=0 at yf= -0.088m
So
The magnitude of Force is 8.58×10⁵N and direction is upwards
The car's acceleration will be 0.575 m/s².The unit of acceleration is m/sec².
<h3>What is acceleration?</h3>The rate of velocity change concerning time is known as acceleration.
Given data;
Initial velocity, u= 0 m/s
Final velocity, v= 4.2 m/s
Time elapsed, t = 7.3 seconds.
To find ;
Acceleration, a
The acceleration when the change in velocity is observed by the formula as:
a= (v-u)/(t)
Substitute the given values:
a= (4.2-0)/(7.3)
a=(4.2)/(7.3)
a= 0.575 m/s²
Hence, the car's acceleration will be 0.575 m/s².
To learn more about acceleration, refer to the link brainly.com/question/2437624
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