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ziang6416's Question
Physics M
Posted 3 months ago
请用中文翻译第三题
用中文解答第二题,第三题,第四题第五题和第六题
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2. Which one of the following is equal to 1 Newton?
A) 1 kg m2/s1 \mathrm{~kg} \mathrm{~m} 2 / \mathrm{s}
B) 1 kg m/s21 \mathrm{~kg} \mathrm{~m} / \mathrm{s}^{2}
C) 1 kg2/s1 \mathrm{~kg}^{2} / \mathrm{s}
D) 1 kg m/s1 \mathrm{~kg} \mathrm{~m} / \mathrm{s}
3.A 20 kg box moving at an initial speed of 10 m/s10 \mathrm{~m} / \mathrm{s} slides 25 m to the right on a horizontal floor before it comes to a complete stop, what is the coefficient of friction between the box and the floor?
(A) 0.1
(B) 0.2
(C) 0.4
(D) 0.6
(E) 0.8
4.A 50.0 N box is at rest on a horizontal surface. The coefficient of static friction between the box and the surface is 0.50 , and the coefficient of kinetic friction is 0.30 . A horizontal 20.0 N force is then exerted on the box. The magnitude of the acceleration of the box is most nearly
(A) 0 m/s20 \mathrm{~m} / \mathrm{s}^{2}
(B) 0.5 m/s20.5 \mathrm{~m} / \mathrm{s}^{2}
(C) 1.0 m/s21.0 \mathrm{~m} / \mathrm{s}^{2}
(D) 4.0 m/s24.0 \mathrm{~m} / \mathrm{s}^{2}
5.Two people, one of mass 100 kg and the other of mass 50 kg , stand facing each other on an ice-covered pond. if the heavier person pushes on the lighter one with a force F , then
(A)The force felt by the heavier person is -0.5 F(B)Theforcefeltbytheheavierpersonis2F
(B)The force felt by the heavier person is -2 F
(C)The magnitude of the acceleration of the lighter person will be 0.5 of the magnitude of the acceleration of the heavier person
(D)The magnitude of the acceleration of the lighter person will be twice the magnitude of the acceleration of the heavier person
6.Three forces act on an object. If the object is in translational equilibrium, which of the following must be true?
I. The vector sum of the three forces must equal zero.
II. The magnitudes of the three forces must be equal
III. All three forces must be parallel
(A) I only
(B)II only
(C)I and III
(D)II and III
斯课堂国际数同 Thinktown Education I New York - Beling - Shanghai - Shenzhen - Hangzhou
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Answer from Sia
Posted 3 months ago
Solution
a
1牛顿的定义是使质量为1千克的物体产生1米每二次方秒的加速度的力。
b
根据牛顿第二定律,F=maF = ma,其中FF是力,mm是质量,aa是加速度。
c
因此,1牛顿等于1kgm/s21 \, \text{kg} \cdot \text{m/s}^2
Answer
B
Key Concept
牛顿第二定律:加速度与作用在物体上的净力成正比,与物体的质量成反比。公式:F=maF = ma
Explanation
1牛顿等于使质量为1千克的物体产生1米每二次方秒的加速度的力。


 第三题
a
使用动能定理:初始动能等于摩擦力做的功。初始动能KEi=12mv2KE_i = \frac{1}{2}mv^2
b
摩擦力做的功W=fkdW = f_k \cdot d,其中fk=μkNf_k = \mu_k N
c
N=mgN = mg,则fk=μkmgf_k = \mu_k mg
d
由动能定理:12mv2=μkmgd\frac{1}{2}mv^2 = \mu_k mgd
e
解出μk\mu_kμk=v22gd\mu_k = \frac{v^2}{2gd}。代入数据:v=10m/sv = 10 \, \text{m/s}d=25md = 25 \, \text{m}g=9.8m/s2g = 9.8 \, \text{m/s}^2
f
μk=(10)229.8250.2\mu_k = \frac{(10)^2}{2 \cdot 9.8 \cdot 25} \approx 0.2
Answer
B
Key Concept
动能定理:初始动能等于摩擦力做的功。公式:KEi=12mv2KE_i = \frac{1}{2}mv^2W=fkdW = f_k \cdot d
Explanation
通过动能定理和摩擦力公式计算出摩擦系数。


 第四题
a
计算静摩擦力:fs=μsNf_s = \mu_s N,其中N=mgN = mg
b
fs=0.5050N=25Nf_s = 0.50 \cdot 50 \, \text{N} = 25 \, \text{N}
c
施加的力20N20 \, \text{N}小于静摩擦力25N25 \, \text{N},因此物体不会移动。
Answer
A
Key Concept
静摩擦力:静摩擦力防止物体相对运动。公式:fs=μsNf_s = \mu_s N
Explanation
施加的力小于静摩擦力,物体不会移动。


 第五题
a
根据牛顿第三定律,作用力和反作用力大小相等,方向相反。
b
设轻者受到的力为FF,重者受到的力为F-F
c
轻者的加速度a1=Fm1a_1 = \frac{F}{m_1},重者的加速度a2=Fm2a_2 = \frac{F}{m_2}
d
因为m1=50kgm_1 = 50 \, \text{kg}m2=100kgm_2 = 100 \, \text{kg},所以a1=2a2a_1 = 2a_2
Answer
D
Key Concept
牛顿第三定律:每一个作用力都有一个大小相等方向相反的反作用力。公式:Faction=FreactionF_{\text{action}} = -F_{\text{reaction}}
Explanation
轻者的加速度是重者加速度的两倍。


 第六题
a
平衡条件:物体处于平衡状态时,合力为零。
b
三个力的矢量和必须为零。
c
力的大小不一定相等,方向也不一定平行。
Answer
A
Key Concept
平衡条件:物体处于平衡状态时,合力为零。公式:F=0\sum F = 0
Explanation
物体处于平衡状态时,三个力的矢量和必须为零。

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