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Zinaida [17]
3 years ago
15

A drag racer starts from the rest and accelerates at 7.4m/s^2.how far will he travel in 2.0 seconds?

Physics
1 answer:
Vesna [10]3 years ago
8 0

Answer:

14.8 m

Explanation:

S= ut + \frac{1}{2}at^{2}

where u = initial velocity

S= (0 \frac{m}{s})(2s) + \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=  \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=14.8 m

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<span>your answer is m = 59 kg</span>
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A 0.59 kg bullfrog is sitting at rest on a level log. how large is the normal force of the log on the bullfrog?
ladessa [460]
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3 years ago
A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work
kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

3 0
3 years ago
Veronica claims that she can throw a dart at a dartboard from a distance of 2 m and hit the 5 cm wide bullseye if she throws the
wariber [46]
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t = 2 m/ 15 m/s = 0.133 s

Then, find the height the dart has fallen from its initial point using this equation: h = 0.5gt²

h = 0.5(9.81 m/s²)(0.133 s)² = 0.0872 m or 8.72 cm

Since the diameter of the bull's eye is only 5 cm, and you started at the same level of the top of the bull's eye, that means the maximum allowance would only be 5 cm. Since it exceeded to 8.72 cm, it means that <em>Veronica will not hit the bull's eye.</em>
3 0
4 years ago
A high-jump athlete leaves the ground, lifting her center of mass 1.8 m and crossing the bar with a horizontal velocity of 1.4 m
Romashka [77]

Answer:

The minimum speed when she leave the ground is 6.10 m/s.

Explanation:

Given that,

Horizontal velocity = 1.4 m/s

Height = 1.8 m

We need to calculate the minimum speed must she leave the ground

Using conservation of energy

K.E+P.E=P.E+K.E

\dfrac{1}{2}mv_{1}^2+0=mgh+\dfrac{1}{2}mv_{2}^2

\dfrac{v_{1}^2}{2}=gh+\dfrac{v_{2}^2}{2}

Put the value into the formula

\dfrac{v_{1}^2}{2}=9.8\times1.8+\dfrac{(1.4)^2}{2}

\dfrac{v_{1}^2}{2}=18.62

v_{1}=\sqrt{2\times18.62}

v_{1}=6.10\ m/s

Hence, The minimum speed when she leave the ground is 6.10 m/s.

6 0
3 years ago
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