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iogann1982 [59]
3 years ago
12

Bằng lý luận và thực tiễn, anh (chị) hãy làm sáng tỏ nhận định: “Tâm lý người là sản phẩm của hoạt động và giao tiếp”, từ đó rút

ra những kết luận cần thiết đối với bản thân.
Physics
1 answer:
Ronch [10]3 years ago
3 0

the awnser is yes because when I did this assignment I got an A

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True or false The momentum of a 100kg object traveling at 3 m/s is 300 N
melomori [17]

Answer:

True

Explanation:

Momentum of an object can be defined as the product of its mass and velocity at which it is travelling. With that in mind, momentum = 3*100=300(kg⋅m/s).

One thing to note is the units mentioned. The SI unit of momentum is kg * m/s as it is the product of mass(kilograms) and velocity(meter per second) and not Newton.

5 0
3 years ago
When the mass of the bottle is 0.125 kg, the average maximum height of the beanbag is m. When the mass of the bottle is 0.250 kg
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Answer:

when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.

when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.

when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.

when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.

Explanation:

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3 years ago
A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak
GarryVolchara [31]

Explanation:

It is given that,

Mass of the box, m = 100 kg          

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.  

From the attached figure, the x and y component of forces is given by :

T_{1x} =-T_1 cos (20)

T_{2x} = T_2 cos (40)

mg_x = 0

T_{1y} = T_1 sin (20)

T_{2y} = T_2 sin (40)

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Let R_x and R_y is the resultant in x and y direction.

R_x=-T_1 cos (20)+T_2 cos (40)+0

R_y=T_1 sin(20)+T_2 sin(40)-mg

As the system is balanced the net force acting on it is 0. So,

-T_1 cos (20)+T_2 cos (40)+0=0.............(1)

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On solving equation (1) and (2) we get:  

T_1=866.86\ N (tension on the left rope)

T_2=1063.36\ N (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.                            

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