Answer:
The magnitude of the force exerted by the locomotive on the caboose is 3.08 N.
Explanation:
Given that,
Mass of a toy, m = 2.3 kg
Mass of caboose, m' = 1 kg
The frictional force of the track on the caboose is 0.48 N backward along the track.
The acceleration of train,
The magnitude of the force exerted by the locomotive on the caboose is given by :
So, the magnitude of the force exerted by the locomotive on the caboose is 3.08 N.
Answer:
There isnt enough in your question to answer the question bro, like we need a picture or something bro.
Explanation:
Answer:
D wavelength
Explanation:
The different wavelengths determine the color.
Answer:
The work done by Joe, W_joe, when he raises the barbell to finish one repetition is 211.98J
The correct option is Wsam <W_Joe.
Explanation:
From the mass of the bar, the acceleration of gravity and the distance that the bar travels in each of them when performing the lifting movement from the chest you can calculate the work done by each one:
W_joe= 31.5kg × 9.81 (m/s^{2} ) × 0.686m = 211.98J
W_sam= 31.5kg × 9.81 (m/s^{2} ) × 0.608m = 187.88J
187.88J < 211.98J ⇒ Wsam <W_Joe
Answer:
the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr
Explanation:
Given:
speed of space vehicle =5000 km/hr
rocket motor speed = 71 km/hr relative to the command module
mass of module = m
mass of motor = 4m
By conservation of linear momentum
Pi = Pf
Pi= initial momentum
Pf= final momentum
Since, the motion is only in single direction
Where M is the mass of the space vehicle which equals the sum of motor's mass and the command's mass, Vi its initial velocity, V_mE is velocity of motor relative to Earth, and V_cE is its velocity of the command relative to Earth.
The velocity of motor relative to Earth equals the velocity of motor relative to command plus the velocity of command relative to Earth.
V_mE = V_mc+V_cE
Where V_mc is the velocity of motor relative to command this yields
substituting the values we get
= 4943.2 Km/hr
the speed of the command module relative to Earth just after the separation = 4943.2 Km/hr