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A 4.5kg ham is thrown into a stationary 15 kg shop cart At what speed will the cart travel if the ham had an initial speed of 2.2 ms?
m = 4.5kg v = 2.2m/s M = 15kg V = ? ---- velocity of ham and cart before: momentum of ham = pv momentum of cart = 0 after: momentum of ham and cart = (m+M)V momentum of ham + momentum of cart = momentum of ham and cart mv + 0 = (m+M)V mv / (m+M) = V (4.5*2.2) / (4.5 + 15) = 0.51 kgm/s
What happen to the speed of a cart when a horse applies more force?
When a horse applies more force to a cart, the speed of the cart will increase. This is due to Newton's second law of motion, which states that the acceleration of an object is directly proportional to the force applied to it. So, the greater the force applied by the horse, the faster the cart will accelerate and increase in speed.
What can you say about the motion of the cart is it in uniform motion or is it accelerating?
If the cart is moving at a constant speed in a straight line, it is in uniform motion. If the cart is changing its speed or direction, it is accelerating.
If a 500N net force is applied to a 38 kg cart for 4 seconds what is the cart's final velocity?
The final velocity of the cart can be calculated using the formula: final velocity = initial velocity + (net force/mass) * time. Assuming the initial velocity is 0 m/s, the final velocity would be: 0 + (500N / 38kg) * 4s = 52.63 m/s.
Which shopping cart is easier to stop physics problem?
In terms of stopping a physics problem involving a shopping cart, it would be easier to stop a cart with a lower mass and slower velocity. This is because the stopping distance is directly related to the mass and velocity of the object. A lighter cart moving at a slower speed will be easier to stop compared to a heavier cart moving at a faster speed.
A 4.5kg ham is thrown into a stationary 15 kg shop cart At what speed will the cart travel if the ham had an initial speed of 2.2 ms?
m = 4.5kg v = 2.2m/s M = 15kg V = ? ---- velocity of ham and cart before: momentum of ham = pv momentum of cart = 0 after: momentum of ham and cart = (m+M)V momentum of ham + momentum of cart = momentum of ham and cart mv + 0 = (m+M)V mv / (m+M) = V (4.5*2.2) / (4.5 + 15) = 0.51 kgm/s
How does the speed of a cart rolling down a ramp change with the mass of the cart?
The speed of a cart rolling down a ramp primarily depends on the angle of the ramp and the acceleration due to gravity, rather than the mass of the cart itself. According to physics, when friction is negligible, all objects accelerate at the same rate regardless of their mass. Therefore, while a heavier cart may have more gravitational force acting on it, it also has more inertia, resulting in the same final speed as a lighter cart at the bottom of the ramp, assuming they start from rest and experience the same conditions.
What happen to the speed of a cart when a horse applies more force?
When a horse applies more force to a cart, the speed of the cart will increase. This is due to Newton's second law of motion, which states that the acceleration of an object is directly proportional to the force applied to it. So, the greater the force applied by the horse, the faster the cart will accelerate and increase in speed.
A cart is 10 cm long It travels through a photogate in 2 seconds what's the cart's speed?
V = 10cm/2s = 5cm/s Therefore, the cart's speed is 5cm/s.
What can you say about the motion of the cart is it in uniform motion or is it accelerating?
If the cart is moving at a constant speed in a straight line, it is in uniform motion. If the cart is changing its speed or direction, it is accelerating.
When a 1.0-kilogram cart moving with a speed of 0.50 meter per second on a horizontal surface collides with a second 1.0-kilogram cart initially at rest the carts lock together What is the speed of?
A very basic model where we can assume there is no momentum loss during the collision... we need to calculate momentum, p.p = mvwhere m = 1 kg (the mass of the cart) and v = 0.5 m/s (velocity of the cart)p = 1 kg . 0.5 m/s = 0.5 kg.m/sAfter the collision the momentum will be the same, but the mass has doubled...so p = mv = 0.5 kg.m/s = (1 kg + 1 kg).vv = 0.5 kg.m/s / 2 kg = 0.25 m/s
If an initially stationary cart rolls down an incline with a measured acceleration of 2.7meters over seconds squared how far does the cart roll in 3point1 seconds?
.281 meters
If a 500N net force is applied to a 38 kg cart for 4 seconds what is the cart's final velocity?
The final velocity of the cart can be calculated using the formula: final velocity = initial velocity + (net force/mass) * time. Assuming the initial velocity is 0 m/s, the final velocity would be: 0 + (500N / 38kg) * 4s = 52.63 m/s.
How do you increase Speed on a Yamaha 48 volt golf cart?
You can increase the speed of an Yamaha 48 volt golf cart by installing a larger motor or more powerful batteries. This will produce more power to the wheels and more speed.
Which shopping cart is easier to stop physics problem?
In terms of stopping a physics problem involving a shopping cart, it would be easier to stop a cart with a lower mass and slower velocity. This is because the stopping distance is directly related to the mass and velocity of the object. A lighter cart moving at a slower speed will be easier to stop compared to a heavier cart moving at a faster speed.
How fast can a 150cc go-cart go?
Max Speed 37-39 mph... that's pretty fast for a go-cart.
As a cart travels around a horizontal circular track does the cart undergo a change in velocity speed inertia or weight?
Cart experiences a change in velocity(which is a vector quantity, not like speed). Cart's velocity on circular track has to be tangent to track at each point and because of that it has to change its direction. Speed may or not remain the same, you can't tell it changes in each possible case. Mass and weight remain the same.
