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Vladimir [108]
3 years ago
10

I have the mass but I can't find (b) or (c)

Physics
1 answer:
Solnce55 [7]3 years ago
5 0

Answer:

i only wrote this for points

Explanation:

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You are carrying a 7.0-kg bag of groceries as you walk at constant velocity along the sidewalk. You walk a distance of 82 meters
MAVERICK [17]

Answer:

0J

Option: B

Explanation:

Work is done when something is moved by the force in the direction of the force. That is the force (e.g., the weight) and the direction the object moves must be aligned for work to be done. In this given condition, the direction is horizontal and the force is downward as its gravity force. That 90° between the two vectors.

The work function is W = m × g ×h × cosθ

\text { Where, in this case theta }=90^{\circ}

Hence,  

Work done = 7 × 9.8 × 1.5 × cos(90)  

Work done = 0 (cos90^{\circ} = 0)

Work done = 0

Therefore work done is 0 J.

7 0
3 years ago
A scooter has wheels with a diameter of 120 mm. What is the angular speed of the wheels when the scooter is moving forward at 6.
nirvana33 [79]

To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

v = r\omega \rightarrow \omega = \frac{v}{r}

Here,

v = Lineal velocity

\omega= Angular velocity

r = Radius

Our values are

v = 6/ms

r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m

Replacing to find the angular velocity we have,

\omega = \frac{6m/s}{0.06m}

\omega = 100rad/s

Convert the units to RPM we have that

\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})

\omega = 955.41rpm

Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm

4 0
4 years ago
A 12.70 g bullet has a muzzle velocity (at the moment it leaves the end of a firearm) of 430 m/s when rifle with a weight of 25.
Norma-Jean [14]

Answer:

2.1844 m/s

Explanation:

The principle of conservation of momentum can be applied here.

when two objects interact, the total momentum remains the same  provided no external forces are acting.

Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\

assume the bullet goes to right side and the gravitational acceleration =10 ms^{-2}

so now the weight of the rifle=\frac{25}{10}  

0=m_{bullet}*v_{bullet}  + m_{rifle}*v_{rifle}  \\\\0=(12.70*10^{-3}) *430ms^{-1} +(\frac{25}{10} )*v_{rifle} \\v_{rifle} =-2.1844ms^{-1}

this is a negative velocity to the right side. that means the rifle recoils to the left side

3 0
3 years ago
A spring is 20cm long is stretched to 25cm by a load of 50N. What will be its length when stretched by 100N. assuming that the e
IgorLugansk [536]

Answer:

Final Length = 30 cm

Explanation:

The relationship between the force applied on a string and its stretching length, within the elastic limit, is given by Hooke's Law:

F = kΔx

where,

F = Force applied

k = spring constant

Δx = change in length of spring

First, we find the spring constant of the spring. For this purpose, we have the following data:

F = 50 N

Δx = change in length = 25 cm  - 20 cm = 5 cm = 0.05 m

Therefore,

50 N = k(0.05 m)

k = 50 N/0.05 m

k = 1000 N/m

Now, we find the change in its length for F = 100 N:

100 N = (1000 N/m)Δx

Δx = (100 N)/(1000 N/m)

Δx = 0.1 m = 10 cm

but,

Δx = Final Length - Initial Length

10 cm = Final Length - 20 cm

Final Length = 10 cm + 20 cm

<u>Final Length = 30 cm</u>

6 0
3 years ago
If you swing an object on the end of a string around a circle, the string pulls on the object to keep it moving in a circle. Wha
emmainna [20.7K]

Answer:

B

Explanation:

4 0
3 years ago
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