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Shalnov [3]
2 years ago
10

Charlie is playing with his daughter Torrey in the snow. She sits on a sled and asks him to slide her across a flat, horizontal

field. Charlie has a choice of (a) pushing her from behind by applying a force downward on her shoulders at ​30^\circ​ below the horizontal (Fig. a), or (b) attaching a rope to the front of the sled and pulling with a force at ​30^\circ​ above the horizontal (Fig. b). Which would be easier for him and why?
Physics
2 answers:
poizon [28]2 years ago
6 0

Answer:

(b) attaching a rope to the front of the sled and pulling with a force at ​30^\circ​ above the horizontal

Explanation:

Remember that it is easier to pull an object than to push it, this is because when you pull you often do it with an angle towards the positive side of the vertical and when you push you often do it towards the negative side of the vertical this increases the friction and the force that is keeping you from being able to move the object.

professor190 [17]2 years ago
3 0

Answer:

b

Explanation:

Pulling up with the rope would decrease the frictional force, pushing down would increase the frictional force.

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A ball was kicked upward at a speed of 64.2 m/s. how fast was the ball going 1.5 seconds later
UNO [17]

Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity.  If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.

                (64.2 m/s)  -  [ (9.8 m/s² ) x (1.5 sec) ] 

            =  (64.2 m/s)  -       [      14.7 m/s      ]

            =             49.5 m/s  .  (upward)

7 0
2 years ago
A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?
OLga [1]

v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

7 0
2 years ago
Read 2 more answers
A 150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.850N)sin
krok68 [10]

Answer:

9.98 m/s

Explanation:

The force acting on the particle is defined by the equation:

F=(0.850) sin (\frac{x}{2.00}) [N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

a=\frac{F}{m}

where

m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

a=\frac{0.850}{0.150}sin (\frac{x}{2.00})=5.67 sin(\frac{x}{2.00}) [m/s^2]

When x = 3.14 m, the acceleration is:

a=5.67 sin(\frac{3.14}{2.00})=5.67 m/s^2

Now we can find the final speed of the particle by using the suvat equation:

v^2-u^2=2ax

where

u = 8.00 m/s is the initial velocity

v is the final velocity

a=5.67 m/s^2

x = 3.14 m is the displacement

Solving for v,

v=\sqrt{u^2+2ax}=\sqrt{8.00^2+2(5.67)(3.14)}=9.98 m/s

And the speed is just the magnitude of the velocity, so 9.98 m/s.

4 0
2 years ago
A physics student walks 100 meters in 80 seconds. The student stops for 30 seconds, and then walks 200 meters farther in 90 seco
Elan Coil [88]

Answer:

B i think is the answer

Explanation:

i feel like it is B because if you put them together and the answer is 1.5 so it is B

8 0
2 years ago
You should have observed that there are some frequencies where the output is stronger than the input. Discuss how that is even p
nydimaria [60]

Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

             X = √ (R² +  (X_{C} -X_{L})² )

where Xc and XL is the capacitive and inductive impedance, respectively

            X_{C} = 1 / wC

           X_{L} = wL

From this expression we can see that for the resonance frequency

           X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

               V = IR

Since the contribution of the two other components is canceled, this occurs for

                X_{C} = X_{L}

                1 / wC = w L

                w = √ 1 / CL

6 0
2 years ago
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