Physics · Work, Energy and Power
500 किलोग्राम वजन वाले एक लिफ्ट को 0.25 मीटर/सेकंड के निरंतर वेग से ऊपर उठाया जाना है। उपयोग की जाने वाली मोटर की न्यूनतम हॉर्सपावर (अश्वशक्ति) क्या होगी? (g = 9.8 मीटर/सेकंड2 लीजिए और मान लीजिए कि कोई घर्षण हानि नहीं है, 1 hp = 746 W)
Verified Solution4 OptionsUpdated: October 2026👁️1 View
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- 12.00 hp
- 21.92 hp
- 31.64 hp✓
- 41.00 hp
सही उत्तर देखें
1.64 hp
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