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madreJ [45]
3 years ago
9

If a statement is true, select true. if it's false, select false. ​

Physics
2 answers:
Elena L [17]3 years ago
7 0

3 is false 2 is true and the rest true

Dimas [21]3 years ago
4 0

Answer;

True, False, False, True, True

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A dumbell has a mass of 95 kg. What force must be applied to accelerate it upward at 2.2 m/s2?
Sveta_85 [38]
A :-) F = ma
Given - m = 95 kg
a = 2.2 m/s^2
Solution -
F = ma
F = 95 x 2.2
F = 209

.:. The force is 209 N
5 0
3 years ago
Two sound waves, A and B, are traveling at the same speed. Wave A has a wavelength of 50 cm and a frequency of 7000 Hz. Wave B h
vekshin1

Answer:

D. 100 cm

Explanation:

The speed of a wave is the wavelength times the frequency.

v = λf

Wave A and B have the same speed, so:

λf = λf

(50 cm) (7000 Hz) = λ (3500 Hz)

λ = 100 cm

6 0
2 years ago
If the half-life of iodine-131 is 8.10 days, how long will it take a 50.00 g sample to decay to 6.25 g?
Sloan [31]

Answer:

your in mr langfords class

Explanation:

bruh moment

6 0
3 years ago
A 44.0 kg uniform rod 4.90 m long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a w
pychu [463]

Answer:

x ≤ 3.6913 m

Explanation:

Given

Mrod = 44.0 kg

L = 4.90 m

Tmax = 1450 N

Mman = 69 kg

A: left end of the rod

B: right end of the rod

x = distance from the left end to the man

If we take torques around the left end as follows

∑τ = 0   ⇒   - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0

⇒   - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0

⇒  -  (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0

⇒ x ≤ 3.6913 m

4 0
3 years ago
The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin
Korolek [52]

Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

K=730 N/m,m=0.053kg, h=1.90m.

v_f^2=v_i^2+2*g*h

v_i^2=2*g*h=2*9.8m/s^2*1.9m

v_i=\sqrt{2*9.8m/s^2*1.9m}=\sqrt{37.24 m^2/s^2}

v_i=6.1 m/s

v_i=V*sin(31)

V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}

V=11.84 m/s

b).

K_k=\frac{1}{2}*K*x^2

No friction on the ball so:

x^2=\frac{2*K_k}{K}

x=\sqrt{\frac{2*0.053kg*9.8m/s^2*1.9m}{730N/m}}

x=\sqrt{2.7x10^{-3}m^2}=0.052m

5 0
3 years ago
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